Given an integer array arr and a 2D integer array queries, where each element of queries represents a query in the form of [s, e], perform the following operation for each query: For each query, increment arr[i] by 1 for all i such that s ≤ i ≤ e. Return the array arr after processing all queries according to the above rule.

Constraints:

The length of arr is between 1 and 1,000 (inclusive). Each element of arr is between 0 and 1,000,000 (inclusive). The length of queries is between 1 and 1,000 (inclusive). For each query [s, e], s and e are between 0 and the length of arr minus 1 (inclusive).

Implementation

vector<int> solution(vector<int> arr, vector<vector<int>> queries) {
    for (int i = 0; i < queries.size(); i++) {
        int start = queries[i][0];
        int end = queries[i][1];
        for (int j = start; j <= end; j++) {
            arr[j]++;
        }
    }
    return arr;
}

instead of creating the output answer, we can just reduce the space, by writing into arr directly.

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation.

The major constraint is to calculate the product of array except itself. If you implement in brute force way, you will get O(n^2). However, this is the constraints! We need to solve this problem for O(n) time complexity with no extra space complexity.

Approach

This 2D structure represents the products of all numbers except the one at each index. The second and third lines show how we can calculate the left and right parts separately.

24, 2, 3, 4
1, 12, 3, 4
1,  2, 8, 4
1,  2, 3, 6

For the input array , the output array represents the prefix products, which are the products of all numbers to the left of each index. This prefix product is calculated by accumulating the numbers to the left at each iteration.

For instance, in the first iteration, there are no numbers to the left, so it’s just 1. In the second iteration, you have 1 to the left, so the output is 1 * 1 = 1. In the third iteration, you have 1 and 2 to the left, so the output becomes 1 * 2 = 2`, and so on.

This process continues until you have the prefix products for all indices. After calculating the prefix products, you can calculate the suffix products and multiply them with the prefix products to get the final result.

Then you do the right, which are going to be the product of suffix. You can do the same thing from right to left.

Implementation

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();

        std::vector<int> output(n, 1);

        // left
        for(int i = 1; i < n; i++) {
            output[i] = output[i - 1] * nums[i - 1];
        }
        
        // right
        int right = 1;
        for (int i = n - 1; i >= 0; i--){
            output[i] *= right;
            right *= nums[i];
        }
        
        return output;
    }
};

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Approach

• the fact that we’re looking for the sum, we just need to subtract the value in the nums, from the summation untill N.

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int N = nums.size();
        int sum = 0;
        int compareSum = 0;
        for (int i = 0; i < N; i++){
            sum += nums[i];
        }
        
        for (int i = 0; i <= N; i++){
            compareSum += i;
        }

        return compareSum - sum;
    }
};

But this is O(N).

Can we make it better ?

Let’s think about this way, we are going to all the power to the sort(), which means the time complexity depends on sort algorithm. Then we can divide three cases:

1. If the first digit is missing, which is mostly likely 0 if the vector is sorted
2. If the last digit is missing, which is mostly likely the max Value of the vector.
3. If the any middle index are missing, starting from 0 to N-1, find the missing number and return the index.

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        if (nums[0] != 0) return 0; // 0 is missing

        if (nums[n-1] != n) return n; // if the last character is missing

        for(int i = 1; i < nums.size(); i++){
            if (nums[i] != i) {
                return i;
            }
        }
        return 0;
    }
};

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Implementation

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* curr = dummyHead;
        int carry = 0;

        while(l1 != nullptr || l2 != nullptr || carry != 0) {
            int x = l1 ? l1->val : 0;
            int y = l2 ? l2->val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            curr->next = new ListNode(sum % 10);
            curr = curr->next;
            l1 = l1 ? l1->next : nullptr;
            l2 = l2 ? l2->next : nullptr;
        }
ㅇㅇ
        ListNode* result = dummyHead->next;
        delete dummyHead;
        return result;
    }
};

Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

Approach

I thought if you put all the numbers in Set, then it will make my life easier, so I created set, then if loop counter is not in, then append to outputs like below

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> output;
        set<int> mySet(nums.begin(), nums.end());
        cout << nums.size() << endl;
        for (int i = 1; i <= nums.size(); i++)
        {
            if (!mySet.contains(i))
                output.push_back(i);
        }
        return output;
    }
};

Then I checked, and the time complexity was indeed O(n), but it took about 83 ms, which wasn’t what I wanted. One thing I was missing was considering the space complexity by switching to a set. However, as I looked at the hint, I realized that using the index itself was the key to improving performance.

One interesting aspect of the implementation below is that the elements at the corresponding index are marked as -1, indicating that the number exists somewhere, but the indices with positive values are the ones where the numbers do not correspond to their indices.

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int n = nums.size();
        vector<int> output;
        for(int i = 0; i < n; i++){
            int index = abs(nums[i]) - 1;
            if (nums[index] > 0){
                nums[index] = -nums[index];
            }
        }

        vector<int> result;
        for(int i = 0; i < n; i++) {
            if (nums[i] > 0)
                result.push_back(i + 1);
        }
        return result;
    }
};

Of course, unordered_map can be also good and fast, but space complexity can be O(n) in that case.

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        unordered_set<int> arr(nums.begin(), nums.end());
        int n = nums.size();
        vector<int> ans;
        for(int i = 1; i <= n; i++) {
            if(arr.find(i) == arr.end()) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

• Reviewing what I studied, how this work will be explained as well.

Union-Find (Disjoint-Set) Concept

The Union-Find algorithm is used to manage disjoint-sets, which are collections of sets that are mutually exclusive, meaning they have no elements in common. This algorithm supports two primary operations: Find: This operation finds the representative element (root) of the set to which a given element belongs. Union: This operation merges two disjoint sets into one.

Disjoint-Set Tree

Disjoint-sets can be represented as a tree structure. Each node represents an element of a set, and child nodes point to their parents. The root node points to itself.

The Union-Find algorithm involves the following key operations:

• Make-Set: Initializes all elements to point to themselves.
• Find: Finds the root (representative element) of the set containing a given element.
• Union: Merges two disjoint sets into one.

int root[max_size];

// Similar to Make-Set function
for (int i = 0; i < max_size; i++) {
    parent[i] = i;
}

int find(int x) { // O(1)
    if (root[x] == x) {
        return x;
    }
    else {
        return find(root[x]);
    }
}

void union(int x, int y) { // O(N)
    x = find(x);
    y = find(y);

    root[y] = x; // set y's root to x's root
}

Optimization on Find / Union.

But there are two issues in the code above. If the depth of tree is 6, and to find that root node, we have to recursively go through 6 steps. Simply, why do we have to take 6 steps to find the root node? can we just make depth 1, but connected to all node! Yes! we can certainly! this method is called path compression.

Another issue would be union. Let’s assume I have trees A and B. A has depth (3), and B has depth (6), Then would it better to merge B into A or A into B? It’s better to merge tree that have lower depth to greater depth. For example, if we put A into B, then the depth of union tree doesn’t change, but if we union B to A, you will get +1 depth in resulting tree. So this can be managed by the concept of rank, which is similar to depth.

The below algorithm has path compression and union by rank applied.

struct UnionFind {
    vector<int> parent; // parent link
    vector<int> rank;   // depth information

    UnionFind(int n): parent(n), rank(n, 0) {
        // set all element pointing to itself.
        for(int i = 0; i < n; i++)
            parent[i] = i; // make-set
    }

    int find(int x) {
        if (parent[x] == x)
        return x;
    }

    int merge(int x, int y) {
        int x_set = find(x);
        int y_set = find(y);

        if (x_set == y_set)
            return x_set

        if (rank[x_set] < rank[y_set]){
            return parent[xset] = yset;
        } else {
            if (rank[x_set] == rank[y_set])
                rank[x_set]++
            return parent[y_set] = x_set;
        }
    }
}

Time Complexity

Let’s look at the time complexity by the operation. makeSet is O(N) when n element is given. Find Operation is similar, but the worst case scenario it’s O(N). Since Union uses Find operation, it takes about O(N). Overall, it becomes O(N) times, but constant time O(a(n)) if path compression & union by rank. (a=> ackerman function).

Resource

• Union-Find

• Reviewing what I studied, how this work will be explained as well.

자, Kruskal Algorithm 은 Union-Find 을 사용해서, 알고르즘을 Process 하며, 결국에는 어떻게 최소의 weight 으로 모든 Vertex 를 연결 할지를 찾기 때문에, greedy choice 를 가져간다. Prim’s 와는 다르게, 전역 Edge 에 대해서 weight 별로 정렬을 한다. 그리고 Union(a, b) or Union(b,a) 를 진행하면서 disjoint-set 들을 만들어간다. 즉 Prim’s 와는 다르게 어떠한 Random 한곳에서 시작하는곳이 아니라 Weight 이 제일 작은 것부터 시작해서, 모든 Edge 별로 disjoint-set 을 만들어가며, Minimum spanning Tree 를 만든다.

생각보다 Union Code 를 활용해서, Path Compression 및 Union-by Rank 를 사용하면, 코드는 간단해진다. 물론 코드가 간단해지지만 Union-Find 가 있다라는 가정하에 이 모든 알고리즘이 쉬워진다고 볼수 있다.

int main()
{
	vector<Edge> edges =
	{
		{ 0, 1, 4.0 },
		{ 0, 7, 9.0 },
		{ 1, 2, 8.0 },
		{ 1, 7, 11.0 },
		{ 2, 3, 7.0 },
		{ 2, 5, 4.0 },
		{ 2, 8, 2.0 },
		{ 3, 4, 9.0 },
		{ 3, 5, 14.0 },
		{ 4, 5, 10.0 },
		{ 5, 6, 2.0 },
		{ 6, 7, 1.0 },
		{ 6, 8, 6.0 },
		{ 7, 8, 7.0 },
	};

	std::sort(edges.begin(), edges.end());
    double mst_wt = 0.0;

    UnionFind uf(9);

    for (auto& e : edges)
    {
	
	    if (uf.Find(e.u) != uf.Find(e.v)) {
		    uf.Union(e.u, e.v);
		    mst_wt += e.weight;
	    }
	    cout << e.u << " - " << e.v << " : " << e.weight << endl;
    }

    cout << mst_wt << endl;
}

Reviewing what I studied, how this work will be explained as well.

Greedy 는 뒷일을 생각하지 않고, 가장 좋아보이는것부터 하나씩 순서대로 처리한다는 의미 (즉 명확한 기준 하나를 정하고 그기준에 따라 미리 정렬을 한다음에 하나씩 처리)! 즉 문제의 크기가 줄어드는 현상이 있다. 문제의 크기가 줄어들기 때문에, 전체의 시간 복잡도가 정렬에 더 치중이 많이 간다. 최적화 문제를 다룬다는 입장에서의 비교! 를 해보자.

Dynamic Programming 안에서는, n 번째 아이템을 넣는다 안넣는다의 차이가 있고, 그거에 따른 n-1 의 선택 을 하는지 안하는지에 따라서 Tree 를 결정할수 있다. Greedy 는 분기를 고려하지 않는다. 가장 좋은 아이템이 있다면, 넣는다, 아니면 안넣는다. 라는 방식으로 분기를 고려할 필요가 없다.

Greedy 의 장점으로서는 1. 구현하기가 쉽고 빠르며, 문제에 따라서 결과가 최적이 아닐 경우가 있다는 소리가 된다.

정리하자면, Greedy Algorithm 에서는 당장 눈에 보이는 최적의 것을 쫓는 알고리즘이라고 할수 있다. 항상 최적은 아니지만, 어느정도의 최적의 근사값? 을 빠르게 구할수 있다라는게 된다.

단순한 문제중 하나는 바로 거스름 돈 문제이다. 항상 적은 양의 화폐를 주는게 제일 편하다. 즉 560 원이라는게 있다면, 10 원짜리 56 개를 주는것이 아닌 500 원짜리 하나, 50 짜리 1개 10 원짜리 1개를 주는것이 총 3개로 편하다. 즉 무조건 더 큰 화폐단위 만큼 거슬러 주는게 좋다라는것으로 생각해서 이 문제는 Greedy 알고리즘으로 풀 수가 있다.

int main() {
    int n, result = 0;
    cin >> n;
    result += n / 500;
    n % = 500;
    result += n / 100;
    n % = 100;
    result += n / 50;
    n %= 50;
    result += n/ 10;
    cout << result;
}

이러한 문제 처럼 주로 무조건 긴대로, 많은대로, 짧은대로 라는 개념이 들어가서, Sort 기법이 들어간다. 대표적인 Kruskal Algorithm 이 될것 같다. 모든 간선을 정렬한 이후 짧은 간선부터 연결하는 MST 가 있을것 같다.

Activity Selection

def activitySelection(start, end):
    ans = 0
    finish = -1

    for i in range(len(start)):
        if start[i] > finish:
            finish = end[i]
            ans += 1
    return ans

def activitySelectionNotSorted(start, end):
    ans = 0
    arr = []
    for i in range(len(start)):
        arr.append((end[i], start[i]))
    
    arr.sort()
    finish = -1

    for i in range(len(start)):
        if start[i] > finish:
            finish = end[i]
            ans += 1
    return ans

if __name__ == "__main__":
    # all sorted by the finished time.
    start = [1, 3, 0, 5, 8, 5]
    end = [2, 4, 6, 7, 9, 9]
    print(activitySelection(start, end))

Resource

Greedy O(n)

• Reviewing what I studied, how this work will be explained as well.

Introduction

As we saw with Dijkstra’s and Bellman-Ford algorithms, we need to check if DP can be applied by identifying optimal substructure and overlapping subproblems. In the context of shortest paths, a portion of a shortest path is itself a shortest path, and overlapping subproblems occur when we reuse partial shortest paths. The approach involves finding three nodes u, v, and k where the path from u to v becomes shorter by going through k.

The key difference between Floyd-Warshall and the other two algorithms is that Floyd-Warshall isn’t fixed to a single starting vertex - it finds the shortest paths between all pairs of vertices. While the algorithm proceeds in stages based on intermediate nodes, it doesn’t require finding the unvisited node with the minimum distance at each step. Due to the graph’s characteristics, Floyd-Warshall can be implemented using an adjacency list, but it’s typically implemented with an adjacency matrix. This makes a bottom-up approach quite feasible.

Recurrence Relation:

Being able to solve a problem with Dynamic Programming means we can define a recurrence relation. At each step, we check if going through a specific node k provides a shorter path. We compare the direct path a->b with the path a->k->b:

Dab = min (Dab, Dak + Dkb) 이런식으로 세울수 있다.

Let’s look at an example using this recurrence relation. The right side shows the graph being updated when k = 1. By applying this process for each node k, we can see the final result in the upper right of the algorithm.

we can infer that the time complexity will be O(n³).

Implementation

코드를 확인해보자…

constexpr double kInf = numeric_limits<double>::infinity();
vector<vector<double>> graph =
{
	{0.0, 3.0, 8.0, kInf, -4.0},
	{kInf, 0.0, kInf, 1.0, 7.0},
	{kInf, 4.0, 0.0, kInf, kInf},
	{2.0, kInf, -5.0, 0.0, kInf},
	{kInf, kInf, kInf, 6.0, 0.0}
};

void FloyedWarshall(const vector<vector<double>>& graph) {
    vector<vector<dobule>> dist = graph;
    int V = graph.size();

    // The graph is already initialized above
    // If not, diagonal values should be 0 and other indices
    // should be filled with appropriate values
    for (int k = 0; k < V; k++) {
        auto dist_k = dist;

        for(int i = 0; i < V; i++) {
            for(int j = 0; j < V; j++) {
                if (dist[i][k] + dist[k][j] < dist[i][j]) {
                    dist_k[i][j] = dist[i][k] + dist[k][j];
                }
            }
        }
    }
}

The implementation is surprisingly simple. We perform updates when necessary. Here’s a Python implementation which might be more straightforward, though it differs in indexing:

INF = int(1e9)

node = int(input())
edge = int(input())
graph = [[INF] * (n+1) for _ in range(n+1)]

for a in range(1, n+1):
    for b in range(1, n+1):
        if a == b:
            graph[a][b] = 0

for _ in range(edge):
    a, b, c = map(int, input().split())
    graph[a][b] = c

for k in range(1, n+1):
    for i in range(1, n+1):
        for j in range(1, n+1):
            graph[i][j] = min(graph[i][j], graph[i][k] + graph[k][j])

To summarize, the Floyd-Warshall algorithm has a time complexity of O(n³).

Resource

• FloydWarshall Algorithm

Cycle detection in a directed graph is a fundamental problem in graph theory with applications in deadlock detection, dependency resolution, and circuit analysis. The algorithm I’ve implemented uses a depth-first search (DFS) approach with a recursive traversal strategy to identify cycles in directed graphs.

Key Concepts

The algorithm relies on three key data structures:

1. A visited flag for each vertex to track which vertices have been explored
2. An on_stack array to track vertices currently in the DFS recursion stack
3. An edge_to array to maintain the DFS tree structure for cycle reconstruction

Implementation

struct Vertex 
{
    Vertex(int v) { index = v; }

    int index = -1;
    bool visited = false;
    vector<Vertex*> out_neighboors;
}

class Graph
{
public:
    Graph(int num_vertices)
    {
        vertices.resize(num_vertices);
        for (int i = 0; i < num_vertices; i++)
            vertices[i] = new Vertex(i);
    }

    ~Graph() 
    {
        for (auto* v : vertices)
            delete v;
    }

    void addEdge(int v, int w)
    {
        verticies[v]->out_neighboors.push_back(vertices[w]);
    }

    void DetectCycle()
    {
        on_stack.resize(vertices.size(), false);
        cycle.clear();
        edge_to.resize(vertices.size(), nullptr);

        for (auto* v : vertices)
            v->visited = false;

        for (auto* v : vertices)
        {
            DetectCycleRecurr(v)
            if (!cycle.empty()) {
                cout << "cycle Detected" << endl;
            }
        }
    }

    void DetectCycleRecurr(Vertex* v)
    {
        v->visited = true;
        on_stack[v->index] = true;

        for(auto* w : v->out_neighboors)
        {
            if (!cycle.empty())
                return;
            else if (!w->visited) 
            {
                edge_to[w->index] = v;
                DetectCycleRecurr(w);
            }
            else if (on_stack[w->index])
            {
                Vertex* x = v;
                cycle.push_back(x);

                while(x->index != w->index) 
                {
                    cycle.push_back(x);
                    x = edge_to[x->index]
                }

                reverse(cycle.begin(), cycle.end())
                return;
            }
        }
    }

private:
    vector<Vertex*> vertices;
    vector<Vertex*> cycle;
    vector<bool> stack;
    vector<Vertex*> edge_to;
}

int main() 
{
    Graph g(3);
    g.addEdge(0, 1);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
}

Time Complexity

Time Complexity: O(V + E) where V is the number of vertices and E is the number of edges. Each vertex and edge is processed exactly once in the worst case

The cycle reconstruction takes at most O(V) time

Space Complexity: O(V)

on_stack, edge_to, and recursion stack each require O(V) space. The cycle array stores at most V vertices