Category: Algorithm

Leetcode 518. Coin Change II

Posted on by NY Comdori

In this post, solution to leetcode problem – 518, coin change II – is explained. Please refer to this link for detail problem description.

This is very similar to combination sum IV problem. The only difference is that the order doesn’t matter and this is essentially a subset problem with duplicates. Let’s see how we can approach the problem.

Brute Force

Exhaustively enumerate all subsets and check if it makes the amount. This will have exponential cost.

Optimized version

We can follow the similar approach as combination sum IV problem. The only difference is that order doesn’t matter and if you blindly add results from previous coins, you will end up counting the duplicate as you see in the diagram below.

In the above diagram, the recurrence relation is simply adding the results from previous amount with all the coin denominations. How can we avoid counting the duplicate? Instead of looking at all the coins for each amount, we can control how many coins each node has. For example, each node is responsible for keeping track of the number of coins for only one coin. Let’s look at the diagram below.

Now, it is not counting the duplicate and you just need to add up all the nodes with valid combinations to get the final result. The recurrence relation would be f(i, a) = sum(f(i-1, a – c*coins[i])) where i is coin index, a is amount, c is number of coins. The range for c is 0 to max coins. How do you calculate max coin? It is simply amount / coins[i]. As there are overlapping subproblems, we can use dynamic programming to solve this. For this, as there are two variables, we would need two dimensional array. row represents coin index and col represents amount. (you can change row/col if you want). What’s the base case? For the first column, when amount = 0, there is only one combination to make the amount – empty subset. For the first row, when no coins are used – I remember I said row is coin index, but since index makes problem slightly more difficult, let’s say row is ith coin that 0th coin means no coin, and the first coin is the first coin in coins – there isn’t any combination possible to make any amount. So everything is 0 except the first column. Although there are more optimizations possible, let’s take a look at the pseudo-code

table: 2d array of row (num coins + 1) col (amount + 1)
initialze first column to 1. everything else is 0

for i in 1 to n: // n is number of coins
    for a in 1 to amount:
        maxCoins = a / coins[i-1]  // note that i is index of table. to access corresponding coin, need to subtract 1
        // check all the possible number of coins can be included
        for c in 0 to maxCoins:
            table[i][a] += table[i-1][a - c*coins[i-1]]
            
return table[n][amount]

Solution is very simple but look at the time complexity. We already have a 2D table and there is an additional for-loop for each cell in the table. Can we do better?

More optimized version

If you closely look at the code, we can eliminate the inner for-loop (c) while achieving the goal. Instead of checking possible number of coins in the for-loop, let’s look at from exclusion/inclusion point of view. At each amount and coin[i], we have a choice – exclude or include. If you exclude, then you check the next coin. If you include, you have another choice – exclude or include the same coin again.

With that, we have a better time complexity. Here is the code written in C++.

int change(int amount, vector<int>& coins) {
    vector<vector<int>> table(coins.size()+1, vector<int>(amount+1, 0));
    for (int i = 0; i <= coins.size(); ++i)
    {
        table[i][0] = 1;
    }

    for (int i = 1; i <= coins.size(); ++i)
    {
        for (int j = 1; j <= amount; ++j)
        {
            // exclude
            table[i][j] = table[i-1][j];

            // include
            if (j >= coins[i-1])
            {
                table[i][j] += table[i][j-coins[i-1]];    
            }
        }
    }
    
    return table[coins.size()][amount];
}

Time complexity=O(n*amount) where n is length of coins and amount is initial amount. Space complexity=(n*amount) because we create 2D array of n*amount. We are very close but there is one more optimization we can achieve – space complexity. If you look at the all the update, exclude code at line 13 is completely unnecessary because after processing the row, next row can start exactly same as previous row. In addition to that, when you look at inclusion code at line 18, you only look at previous column because you are including the same coin!

It means that we only need a 1D array. Here is the C++ code.

int change(int amount, vector<int>& coins) {
    vector<int> table(amount+1, 0);
    table[0] = 1;

    for (int i = 1; i <= coins.size(); ++i)
    {
        for (int j = 1; j <= amount; ++j)
        {
            if (j >= coins[i-1])
            {
                table[j] += table[j-coins[i-1]];
            }
        }
    }
    
    return table[amount];
}

Time complexity is unchanged because you have to iterate all the coins with all the amount. However, we reduced the space complexity to O(amount).

Leetcode 377. Combination Sum IV

Posted on by NY Comdori

This post explains how to solve leetcode 377. combination sum IV problem. It starts with brute force idea and optimize to efficiently solve the problem. Please refer to this link for problem detail.

In this problem, you have two arguments – target and nums array. You are supposed to find the number of combinations to make the target amount only using numbers in nums array. In this problem, the order matters as you see in the example.

Let’s think about brute force idea – enumerate all the possible subsets of nums array. For each subset, you also need to check how many of each number can make the target. This is very expensive with exponential time complexity.

Can we do better? Let’s suppose nums = [1,2,3] and target is 10. Now, let’s also suppose that we have collected a number of combinations right before 10. In other words, we have the result at 9, 8, 7 which are (10 – 1), (10 – 2), (10 – 3). Since the order matters, we need to count all three cases. Then, the answer would be result(10 – 1) + result(10 – 2) + result(10 – 3). We can recursively derive the same relationship for 9, 8, 7 values. Let’s take a look at the below diagram.

We just need to recursively look for children until the base case is reached, which is 0. This is better than enumerating all the subsets and finding combinations. However, there is still room to improve because there are overlapping subproblems. This overlapping subproblem is like calculating fibonacci number recursively. We prevent calculating the same subproblem by storing the result. If subproblem is found, then we just need to reuse that one. Ultimately this becomes dynamic programming.

There are two ways – bottom up, top down – to solve the problem. In this post, I will just explain bottom up approach.

Let’s define f(t) as “number of combinations based on number in nums at target amount t”

Subsequently, the recurrence relation would be f(t) = sum(f(t – nums[i])), where i is 0..nums.length() (index). If t – nums[i] < 0, then it would be 0 as you can’t have negative value.

With the recurrence relation clearly defined, the code becomes very simple. The solution is based on C++.

Base case is t=0. There is only 1 way to make amount 0 which is empty subset. Based on the base case, we need to build up from the bottom until reach the target amount.

int combinationSum4(vector<int>& nums, int target) {
    vector<unsigned int> table(target+1);
    table[0] = 1;
    for (int i = 0; i <= target; ++i)
    {
        for (int num : nums)
        {
            if (i >= num)
            {
                table[i] += table[i-num];
            }
        }
    }

    return table[target];
}

The time complexity is O(nt) where n is length of nums and t is target amount. One gotcha is that if t is really large (ex. > 2^n) this essentially becomes same as brute force solution.

Maximum Repeating Substring

Posted on by NY Comdori

This is an easy level coding interview problem from leetcode. The problem asks you to find the maximum number of repeating string which is given. Please refer to the problem description for more detail. For the solution, you need to iterate the sequence. At each index, you check if the substring matches the word. If it matches, then continue to the next index to find the end of the repeated word. If it doesn’t match, then just move on to the next index.

I provided solutions in C++ and Python3. You can try iterative solution but I provided recursive solutions here. For recursive solutions, it is following the same principles. But note that I keep track of the start index of the comparison because that will be used when the substring doesn’t match. 

void maxRepeating(const string &seq, int currIdx, int matchIdx, const string &word, int numRepeat, int &maxRepeat)
{
    // this is base case. make sure you get the max repeat.
    if (seq.size() - matchIdx < word.size())
    {
        maxRepeat = max(maxRepeat, numRepeat);
    }
    // word doesn't match. get the max repeat and then continue comparing the string from the currIdx which is the start of the matching
    else if (seq.substr(matchIdx, word.size()) != word)
    {
        maxRepeat = max(maxRepeat, numRepeat);
        maxRepeating(seq, currIdx + 1, currIdx + 1, word, 0, maxRepeat);
    }
    // word matches. keep checking. make sure currIdx is passed to note the start of the pattern
    else
    {
        maxRepeating(seq, currIdx, matchIdx + word.size(), word, numRepeat + 1, maxRepeat);
    }
}

int maxRepeating(string sequence, string word) {
    int maxRepeat = 0;
    maxRepeating(sequence, 0, 0, word, 0, maxRepeat);
    return maxRepeat;
}
class Solution:
    result = 0
    
    def max_repeating(self, sequence, curr_idx, match_idx, word, num_repeat):
        compare = sequence[match_idx : match_idx + len(word)]
        if len(compare) < len(word):
            self.result = max(self.result, num_repeat)
            return
        elif compare != word:
            self.result = max(self.result, num_repeat)
            self.max_repeating(sequence, curr_idx + 1, curr_idx + 1, word, 0)
            return 
        
        self.max_repeating(sequence, curr_idx, match_idx + len(word), word, num_repeat + 1)
        
    
    def maxRepeating(self, sequence: str, word: str) -> int:
        self.max_repeating(sequence, 0, 0, word, 0)
        return self.result

Longer Contiguous Segments of Ones than Zeros

Posted on by NY Comdori

This is an easy level leetcode problem. Please refer to this link for the detail of the problem description.

Basically, the idea is to find the longest consecutive 1’s and 0’s and return True if 1’s > 0’s.

There are a couple of ways to achieve the problem.

The first one is a pythonic way. Since the string is a binary string that only has 1 and 0, you can split the string based on them. After the split, it’s really to find the longest string in the array. But this solution is only applicable to python.

class Solution:
    def checkZeroOnes(self, s: str) -> bool:
        return max((len(digit) for digit in s.split('0'))) > max((len(digit) for digit in s.split('1')))

Another solution is to explicitly iterate the string and count num 1’s and 0’s which are applicable to all other languages. You can do it in one pass as you see the solution below.

class Solution:
    def checkZeroOnes(self, s: str) -> bool:
        ones_start_idx = 0
        num_ones = 0
        zeros_start_idx = 0
        num_zeros = 0
        for idx, digit in enumerate(s):
            if digit == '0':
                ones_start_idx = idx + 1
                num_zeros = max(num_zeros, idx - zeros_start_idx + 1)    
            else:
                zeros_start_idx = idx + 1
                num_ones = max(num_ones, idx - ones_start_idx + 1)
            
            
        return num_ones > num_zeros

Delete Characters to Make Fancy String

Posted on by NY Comdori

This is an easy level leetcode problem, which you can use stack to solve.

Please refer to this link for more detail of the problem.

Essentially, you cannot accept 3 or more consecutive duplicate letters. There could be many ways but using a stack seems to be the most elegant approach. Basically, you accumulate each letter if it doesn’t match with the top two entries in the stack or stack has less than 2 elements. This way, you are only accumulating letters that are not duplicate of 3 or more consecutive characters.

class Solution:
    def makeFancyString(self, s: str) -> str:
        fancy = []
        for letter in s:
            if len(fancy) < 2 or fancy[-1] != letter or fancy[-2] != letter:
                fancy.append(letter)
            
        return "".join(fancy)

Check Whether Two Strings are Almost Equivalent

Posted on by NY Comdori

This is a leetcode interview problem that requires a hash table with strings. Please refer to this link for more detail of the problem description.

Let me first show you the solution code.

class Solution:
    def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
        def default_val():
            return list((0, 0))
        count = defaultdict(default_val)
        for letter in word1:
            count[letter][0] += 1
        
        for letter in word2:
            count[letter][1] += 1
            
        for letter, freq in count.items():
            if abs(freq[0] - freq[1]) > 3:
                return False
            
        return True

A hash table is necessary in order to keep track of the frequencies of the letter in the word. Note that I used a list of which size is 2 – the first index is frequencies for word1 and the second index is frequencies for word2.

Once you collected the frequencies, you just need to check if the frequency difference is greater than 3.

Intro to dynamic programming

Posted on by NY Comdori

This article is a simple intro to dynamic programming. Dynamic programming is typically considered the hardest algorithm to work with. However, there is a systematic way to solve the problem and we just need to practice based on that method.

Dynamic programming is best suited for the situation that needs to eliminate duplicate computations by caching it. Now let’s take a look at what steps are necessary to solve the problem.

Steps to solve DP

  • Define the state
  • List out all the state transitions
  • Implement a recursive solution
  • Memoize (top-down)
  • Make it bottom-up

Define the state

What are states in a dynamic programming context?

States are a set of parameters that define the system. Basically, we are trying to find the least number of parameters that change the state of the system. These are the inputs you are passing into the recursive function. Most of the time, it is just one or two parameters that define the state of the system.

After you define the state, you need to define the cost function and its return. It is the one we are trying to optimize for.

Example – knapsack problem

Provided a set of items, each with a weight and a value, determine the combination of each item to include in a bag so that the total weight is less than or equal to a given limit while having the largest total value possible.

states:
1. W – Available capacity of the back (limit)
2. i – Index of the item being considered

Cost function: knapsack(W, i), returns the largest total value that is less than or equal to the limit

Define the state transitions and optimal choice

At this step, you need to find out how the recurrence relation would be and what parameters you will pass to the recursive function. And how do we combine the result?

In order to find out the recurrence relation, you first need to identify the base cases.
1. very last state, where there are no more items to process.
2. some parameters are 0 or reached a final value that we are not able to proceed further.

After identifying the base cases, we need to define transitions. We need to identify all the valid candidates and try each of them one at a time (backtracking). Then, change parameters based on selected candidates and call recursive function. After trying all the candidates, the results need to be combined to find the optimal value we want to return.

Example – knapsack problem

Base cases:
1. If W = 0, it means the knapsack is full. Don’t have enough space for additional items.
2. If i = -1 (starting from last to 0), we processed all the items and nothing is left.
In those cases, the function returns 0

Transitions:
At each item, we have two decisions to make.
1. Take i-th item in the knapsack, then W will become W – weight[i]. It means weight[i] <= W must be true.
2. Skip the i-th item, and W remains the same.

This is the recurrence relation you see.

                                knapsack(W, i)
                                /           \
value[i] + knapsack(W - weight[i], i-1)      knapsack(W, i-1)
    
then, optimal choice would be:
max(value[i] + knapsack(W - weight[i], i-1), knapsack(W, i-1))


Recurrence relation
Base cases:
knapsack(0, i) = 0
knapsack(W, -1) = 0

if weight[i] <= W
    knapsack(W, i) = max(value[i] + knapsack(W - weight[i], i - 1), knapsack(W, i - 1))
else
    knapsack(W, i) = knapsack(W, i - 1)

Implement a recursive solution

This is python version but should be easy enough to understand.

def knapsack(W, i, weights, values):
    if W == 0 or i == -1:
        return 0
    if weights[i] <= W:
        return max(
            values[i] + knapsack(W - weights[i], i - 1, weights, values),
            knapsack(W, i - 1, weights, values)
        )
    else:
        return knapsack(W, i - 1, weights, values)

Memoize (top-down)

The naive recursive solution is slow – O(2^n) – and we can improve the performance by using memoization. How do we use it?
We cache the results of subproblems to avoid computing duplicate subproblems.

If there is only one parameter, we can use a 1D array and use the value of the parameter as the index to store the result.
If there are two parameters, then we use a 2D array – matrix.
Please note to initialize the array with default values.

def knapsack(W, i, weights, values, dp):
    if W == 0 or i == -1:
        return 0
    # check the dp to avoid duplicate computation
    if dp[W][i] != -1:
        return dp[W][i]
        
    # not computed yet
    result = None
    if weights[i] <= W:
        result = max(
            values[i] + knapsack(W - weights[i], i - 1, weights, values, dp),
            knapsack(W, i - 1, weights, values, dp)
        )
    else:
        result = knapsack(W, i - 1, weights, values, dp)
    
    dp[W][i] = result
    return result

Now you completed a dp problem with top-down approach.

Bottom up

If we already completed the problem, why do we even need to solve it bottom up?
There can be a couple of reasons to do that.
1. You might get stack overflow error due to many recursive calls
2. Bottom up is faster than top-down

In bottom-up approach, you have to think in the reverse direction. You need to start at the leaf/bottom of the tree, solve all the problems starting from the leaves and make the way up. It means we will start with the base case and compute upward to final state. We use one for loop for each parameter and fill all the values in the table by solving smaller problems into big ones.

Now, we need to decide which problem to solve first. It actually depends on how the parameters change in recurrence relation. If the parameter decreases in the subproblem then we start from 0 and go all the way up to the maximum value of the parameter.

Base case
if i = 0, return 0
if w = 0, return 0

Bottom up equation
dp[w][i] = max(values[i] + dp[w-weight[i]][i], dp[w][i-1])
def knapsack(W, weights, values):
    dp = [[0 for i in range(0, len(weights) + 1)] for j in range(0, W + 1)]
    for i in range(1, len(weights) + 1):
        for w in range(0, W + 1):
            if weights[i-1] <= w:
                dp[w][i] = max(dp[w]][i-1], dp[w-weights[i-1]][i-1] + values[i-1])
            else:
                dp[w][i] = dp[w][i-1]
    return dp[W][len(weights)]

At first glance, bottom-up solution doesn’t look intuitive at all. However, after solving top-down and defining proper equation it totally makes sense. Dynamic programming is hard but you will definitely get better by practicing!

Dijkstra’s Algorithm Shortest Path

Posted on by NY Comdori

Dijkstra’s algorithm is the most popular algorithm to find the shortest paths from a certain vertex in a weighted graph.
In fact, the algorithm will find the shortest paths to every vertex from the start vertex.
The algorithm takes both greedy and dynamic programming approach that it will try to find local optimum paths first and use them to find the shortest paths.
Let’s take a look at how it works!

How does Dijkstra’s algorithm work?

Let’s say we need to find the shortest path from node A to node E.

dijkstra's algorithm shortest path

There are five nodes and each edge is weighted with the cost to traverse from a node to a node.
For example, it costs 5 to traverse from node A to B.

What would be the most efficient way to find out the shortest path?
In order to find the shortest path from node A to E, we need to find out what is the shortest path to nodes that are right before E – A, B, D.
Can you see where we are going?
In order to find out the shortest path to nodes B, D (I skipped A since it’s the start node), we need to find the shortest path to nodes right before – A, C.

Let’s take a look at each step.

dijkstra's algorithm shortest path

From node A, we can reach to nodes B, C, D, and E.
But let’s start with the nodes B and C first.
What is the shortest path from A to B? There is only one path and it costs 5.
What is the shortest path from A to C? There is only one path and it costs 1.
The nodes B and C are very simple but it’s very important because this will be stepping stone to find the shortest paths to further nodes.

dijkstra's algorithm shortest path

Now, based on the shortest paths to the node C, we can find out the shortest path from the node A to the node D.
There are two paths – A to D directly and A to C to D.
As we can see the shortest path from A to D is A -> C -> D with the total cost 3.

Finally, we have found all the shortest paths from A to B, C, and D which are the nodes right before E.
Based on the shortest paths we found so far, we just need to find which is the cheapest way to reach to E.

dijkstra's algorithm shortest path

There are a total of four possible routes from A to E.

  • A -> E : 10
  • A -> B -> E : 11
  • A -> D -> E : 9
  • A -> C -> D -> E : 8

The shortest path from A to E is A -> C -> D -> E with the cost 8.

The graph itself is pretty simple.
However, the steps we took to find the shortest path is like this.
It really finds the shortest paths to adjacent nodes of A then builds the path based on previous best paths each round.

  • Start from the start node A
  • Just like breadth-first search, find shortest paths to adjacent nodes which are B, C, D, and E.
  • Make sure you keep track of those records.
  • From nodes B, C, and D, find the shortest path to the node E.

Code Example

This code example has the implementation of Dijkstra’s algorithm which uses the graph in the above example.

#include <iostream>
#include <vector>
#include <limits>
#include <algorithm>

using namespace std;

struct Edge
{
    int destNodeId;
    int weight;
    
    Edge(int _destNodeId, int _weight) : 
        destNodeId(_destNodeId), weight(_weight)
    {}
};

struct Node
{
    // node id which is index of vector in this case
    int nodeId;
    
    // set of edges belong to this node
    vector<Edge> edges;

    // indicates if the node is processed - the shortest path found.    
    bool processed;
    
    Node(int _nodeId, vector<Edge> _edges) : 
        nodeId(_nodeId), edges(_edges), processed(false)
    {}
};

struct Graph
{
    // list of nodes in this graph
    vector<Node> nodes;
    
    // indicates the shortest paths up to this node so far.
    vector<int> shortestDistSoFar;
    
    // indicates parent node id for each node id, which is represented by index
    vector<int> parentNodes;
};

void initializeGraph(Graph &graph)
{
    // initialize node A
    vector<Edge> edges;
    edges.emplace_back(1, 5);
    edges.emplace_back(2, 1);
    edges.emplace_back(3, 4);
    edges.emplace_back(4, 10);
    
    graph.nodes.emplace_back(0, edges);
    
    // initialize node B
    edges.clear();
    edges.emplace_back(4, 6);
    
    graph.nodes.emplace_back(1, edges);
    
    // initialize node C
    edges.clear();
    edges.emplace_back(3, 2);
    
    graph.nodes.emplace_back(2, edges);
    
    // initialize node D
    edges.clear();
    edges.emplace_back(4, 5);
    
    graph.nodes.emplace_back(3, edges);
    
    // initialize node E
    edges.clear();
    graph.nodes.emplace_back(4, edges);
    
    graph.shortestDistSoFar.resize(5, numeric_limits<int>::max());
    graph.parentNodes.resize(5, -1);
}

vector<Node> dijkstra(Graph &graph, int startNodeId)
{
    int currNodeId = startNodeId;
    // distance from A to A is 0
    graph.shortestDistSoFar[currNodeId] = 0;
    // A doesn't have the parent node
    graph.parentNodes[currNodeId] = 0;

    while (!graph.nodes[currNodeId].processed)
    {
        graph.nodes[currNodeId].processed = true;
        
        // take a look at adjacent nodes
        for_each(
            graph.nodes[currNodeId].edges.begin(),
            graph.nodes[currNodeId].edges.end(),
            [&graph, currNodeId](const Edge &destNodeEdge)
            {
                // an adjacent node from current node being processed
                int destNodeId = destNodeEdge.destNodeId;
                
                // weight of the edge
                int weight = destNodeEdge.weight;
                
                // if this is shorter than the record, update it
                if (graph.shortestDistSoFar[currNodeId] + weight < graph.shortestDistSoFar[destNodeId])
                {
                    graph.parentNodes[destNodeId] = currNodeId;
                    graph.shortestDistSoFar[destNodeId] = graph.shortestDistSoFar[currNodeId] + weight;
                }
            });
            
        // find next node to process
        // need to process shortest distance first
        int minDistSoFar = numeric_limits<int>::max();
        for_each(
            graph.nodes.begin(),
            graph.nodes.end(),
            [&currNodeId, &graph, &minDistSoFar](const Node &node)
            {
                if (!node.processed && graph.shortestDistSoFar[node.nodeId] < minDistSoFar)
                {
                    minDistSoFar = graph.shortestDistSoFar[node.nodeId];
                    currNodeId = node.nodeId;
                }
            });
    }
    
    vector<Node> shortestPath;
    
    // need to find shortest path by recursively tracking parent node
    
    currNodeId = 4;
    while (graph.parentNodes[currNodeId] != currNodeId)
    {
        shortestPath.push_back(graph.nodes[currNodeId]);
        currNodeId = graph.parentNodes[currNodeId];
    }
    
    // make sure A is in the path
    shortestPath.push_back(graph.nodes[0]);
    reverse(shortestPath.begin(), shortestPath.end());
    return shortestPath;
}

int main()
{
    Graph graph;
    initializeGraph(graph);
    
    // dijkstra from A of which node id is 0
    vector<Node> shortestPath = dijkstra(graph, 0);
    
    for_each(
        shortestPath.begin(),
        shortestPath.end(),
        [](const Node &node)
        {
            switch (node.nodeId)
            {
            case 0:
                cout << "A -> ";
                break;
            case 1:
                cout << "B -> ";
                break;
            case 2:
                cout << "C -> ";
                break;
            case 3:
                cout << "D -> ";
                break;
            case 4:
                cout << "E ";
                break;
            }
        });
    cout << endl;
    return 0;
}

Performance

Time complexity: O(n^2)
It’s because the algorithm needs to visit each node to find the shortest path.
(You can simply find out there is nested loop)

Conclusion

The algorithm works correctly as long as there aren’t any negative weight edges.
The reason is that the huge negative edge may change the shortest path record to certain edge which was already processed.
But I assume there won’t be many cases that graph will have negative edges.
Thank you for reading the post and please let me know if you have any questions/suggestions.

How to detect a loop (cycle) in a linked list

Posted on by NY Comdori

Traversing linked list is easy but what would happen if there is a cycle in the linked list?
It will just fall into an infinity loop if traversing is implemented without a cycle prevention code.
Thankfully, there is already a known algorithm for this kind of problem.

How to detect if a cycle exists?

Suppose there is a linked list that has a cycle like below.
Now the question is how to detect the cycle.

linked list cycle example

Let’s first think about what would happen if there is a cycle.

Iterating linked list typically happens in a while loop like these steps.

  1. Check if the node is valid (not null)
  2. Process node
  3. Move on to next node and do #1,2

Normally iteration will be done once it reaches the end of the list of which next node is null.
However, it will fall into an infinite loop if there is a cycle since there won’t be any null node.

Then, the key is that we shouldn’t visit a node if it was already visited.
How do we detect visited nodes?
The easy answer is to have a cache (map, array or whatever) that keeps track of visited nodes.

So there will be an additional step for cycle detection

  1. Check if the node is valid (not null)
  2. Process node
  3. Move on to next node
  4. Check if this node is already visited. If yes then there is a cycle.

What is the time complexity of this problem? O(n). We only need to iterate once.
What is the space complexity of this problem? O(n). Potentially we need to keep track of all visited nodes.

Can we do better?

We cannot do better on time complexity because you need to iterate the list at least once.
However, there is definitely a way to improve space complexity.
In fact, we can solve this problem with O(1) space complexity.

How can we do that?
Think about two people running on a circular track.
And let’s say one person is able to run twice faster than the other.
Then what would happen in that case?
The faster runner will catch the slow runner at the start point after 1 lap of slow person since the faster runner is able to run track twice while the slow runner finishes 1 lap.
We can apply exact same principle to detect a cycle in a linked list.
If we have two pointers that one pointer iterates twice faster than the other faster pointer will catch up slower pointer once it completes 1 lap.

How to find out where the cycle starts?

I think some visualization will do a much better job than a verbal explanation.
So let’s take a look.

linked list cycle example

The cycle may start after a few nodes.
But as you can see in the picture, the number of nodes before cycle essentially means the number of nodes before cycle start node where two pointers begin iteration.

The above picture shows you that node 15 will be the location where two pointers will meet.
Then, all we need to do is have another pointer at the start of the linked list and iterate both pointer 1 (slow pointer) and 3rd pointer until they meet which is the cycle node!

Code Example

struct Node 
{
    int val;
    Node *next;
    Node(int x) : val(x), next(0) {}
};

Node *detectCycle(Node *head) 
{
    if (!head)
    {
        return 0;
    }

    Node *iter1 = head->next;
    Node *iter2 = head->next ? head->next->next : 0;

    while (iter1 && iter2)
    {
        // cycle found!
        if (iter1 == iter2)
        {
            break;
        }

        iter1 = iter1->next;
        iter2 = iter2->next ? iter2->next->next : 0;
    }

    // no cycle
    if (!iter1 || !iter2)
    {
        return 0;
    }

    Node *cycleNode = head;

    while (iter1 != cycleNode)
    {
        iter1 = iter1->next;
        cycleNode = cycleNode->next;
    }

    return cycleNode;
}

Intro to Topological Sort

Posted on by NY Comdori

Topological sort is an important algorithm in DAG – directed acyclic graphs.

It orders each vertex in line from left to right such that left most vertex could be considered as the highest level or root or entrance of the graph and rightmost as the lowest of the graph.

In the end, topological sort tells you the order of vertices you need to go through from one node to the other without missing any middle vertices.

There could be multiple topological sorts in the same graph.

Let’s take a look at an example for a better understanding

Course Prerequisite Example

topological sort

The course prerequisite is a great example of this kind of problem.

Let’s say there are 4 courses A, B, C, D and here are some prerequisites for each course.

  1. Course C requires A, B, D
  2. Course B requires A
  3. CourseD requires A, B
  4. A is a fundamental course you need to complete before any courses

With that in mind, in order to reach from A to C, we need to follow these steps.

  1. We see a path to node C directly from A but we also see B and D
  2. Visit B
  3. From B we see a path to C and D
  4. Visit D
  5. Now the order of node visit is A, B, D, C which is the correct order for course completion

However, the above steps are not complete yet for topological sort algorithm since some intuition plays a role there.
Here is the correct algorithm.

Topological Sort Algorithm

  1. Visit a node and mark it visited (but not processed yet)
  2. Search any nodes connected to the node from #1 and visit only if it’s not visited yet
  3. If you see any visited but not processed node then there is a cycle and you cannot find topological sort. (It will be an infinite loop!)
  4. Mark the node processed and push it into stack
  5. After all the nodes are processed pop each value from stack and pop order is the correct order for topological sort

Code Example

enum Status
{
    UNDISCOVERED,
    DISCOVERED,
    PROCESSED
};

struct Node
{
    Status      status;
    vector<int> destNodeIds;
};

// return false if there is a cycle
bool findOrderHelper(int currNodeId, unordered_map<int, Node> &nodes, vector<int> &order)
{
    auto it = nodes.find(currNodeId);
    it->second.status = DISCOVERED;

    bool canPublish = true;

    for_each(
        it->second.destNodeIds.begin(),
        it->second.destNodeIds.end(),
        [&nodes, &canPublish, &order](int destNodeId)
        {
            auto it = nodes.find(destNodeId);
            if (it->second.status == UNDISCOVERED)
            {
                canPublish &= findOrderHelper(destNodeId, nodes, order);
            }
            else if (it->second.status == DISCOVERED)
            {
                canPublish = false;
            }
        });

    it->second.status = PROCESSED;
    order.push_back(currNodeId);
    return canPublish;
}

vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) 
{
    vector<int> order;

    // nodes is a graph constructed from prerequisites vector
    unordered_map<int, Node> nodes;
    for_each(
        prerequisites.begin(),
        prerequisites.end(),
        [&nodes](const vector<int> &prerequisite)
        {
            auto it = nodes.insert(make_pair(prerequisite.back(), Node()));
            it.first->second.destNodeIds.push_back(prerequisite.front());

            nodes.insert(make_pair(prerequisite.front(), Node()));
        });

    // now iterate nodes (graph) to find topological sort
    bool canPublish = true;
    for_each(
        nodes.begin(),
        nodes.end(),
        [&nodes, &order, &canPublish](const pair<int, Node> &nodeInfo)
        {
            if (nodeInfo.second.status == UNDISCOVERED)
            {
                canPublish &= findOrderHelper(nodeInfo.first, nodes, order);
            }
        });

    if (!canPublish)
    {
        order.clear();
    }
    else if (order.size() != numCourses)
    {
        // courses don't have many prerequisites
        for (int i = 0; i < numCourses; ++i)
        {
            auto it = find(order.begin(), order.end(), i);
            if (it == order.end())
            {
                order.push_back(i);
            }
        }
    }

    reverse(order.begin(), order.end());

    return order;
}