#include <iostream>
#include <vector>
#include <unordered_map>
#include <cmath>
#include <algorithm>
using namespace std;
class UnionFind {
public:
    UnionFind(int n) : parent(n), size(n, 1) {
        for (int i = 0; i < n; ++i) {
            parent[i] = i;
        }
    }
    int find(int x) {
        if (x != parent[x]) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }
    void unionSet(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY) {
            if (size[rootX] < size[rootY]) {
                swap(rootX, rootY);
            }
            parent[rootY] = rootX;
            size[rootX] += size[rootY];
        }
    }
    int getSize(int x) {
        return size[find(x)];
    }
private:
    vector<int> parent;
    vector<int> size;
};
vector<int> primeFactors(int x) {
    vector<int> factors;
    for (int i = 2; i <= sqrt(x); ++i) {
        if (x % i == 0) {
            factors.push_back(i);
            while (x % i == 0) {
                x /= i;
            }
        }
    }
    if (x > 1) {
        factors.push_back(x);
    }
    return factors;
}
int largestMagicalGroup(int N, vector<int>& A) {
    UnionFind uf(*max_element(A.begin(), A.end()) + 1);
    unordered_map<int, int> value_to_index;
    for (int i = 0; i < N; ++i) {
        value_to_index[A[i]] = i;
        for (int prime : primeFactors(A[i])) {
            uf.unionSet(A[i], prime);
        }
    }
    unordered_map<int, int> group_sizes;
    int max_group_size = 1;
    for (int i = 0; i < N; ++i) {
        int root = uf.find(A[i]);
        group_sizes[root]++;
        max_group_size = max(max_group_size, group_sizes[root]);
    }
    return max_group_size;
}

Magical Group ✅

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How many people got this

T* extendT(T* node) {
    if (!node) return node;

    if (!node->l && !node->r) {
        node->l = new T(node->x,nullptr,nullptr);
        node->r = new T(node->x,nullptr,nullptr);
    } else {
        extendT(node->l);
        extendT(node->r);
    }
    return node;
}

Microsoft ✅

#include <iostream>
#include <vector>
using namespace std;
int solution(vector<string> &A) {
    int RR = 0, GG = 0, RG = 0, GR = 0;
    for (string &s : A) {
        if (s == "RR") RR++;
        else if (s == "GG") GG++;
        else if (s == "RG") RG++;
        else GR++;
    }
   
    if (RG || GR)
        return RR + GG + 2 * min(RG, GR) + (RG != GR);
    else
        return max(RR, GG);
}

Microsoft ✅

#include <iostream>
#include <vector>
#include <unordered_set>

Vector<int> getGoodElements(vector<int>a){
    unordered_set<int> set;
    vector<int>ans;
   
   
    for (int it : a) {
        unordered_set<int> tmp(set);
        for (int itt : set) {
            if (itt < it) {
                tmp.insert(it | itt);
            }
        }
        set = tmp;
        set.insert(it);
    }
    ans.push_back(0);
    for(int i:set) ans.push_back(i);
    return ans;
   
   
    return 0;
}

Citadel ✅

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ll solve(vector<int>& heights) {
    int n = heights.size();
    vector<ll> left(n), right(n, n);
    stack<int> s;
    for (int i = 0; i < n; ++i) {
        while (!s.empty() && heights[s.top()] <= heights[i]) {
            right[s.top()] = i;
            s.pop();
        }
        left[i] = s.empty() ? -1 : s.top();
        s.push(i);
    }
    ll tot = 0;
    for (int i = 0; i < n; ++i) {
        tot += (i - left[i]) + (right[i] - i) - 1;
    }
    return tot;
}.

// CALCULATE REGION
MSCI ✅

#include <iostream>
#include <vector>
using namespace std;
const int MOD = 10000000;
int solve(int N, const string& a) {
    vector<vector<int>> dp(N, vector<int>(3, 0));
    if (a[0] != 'R') dp[0][0] = 1;
    if (a[0] != 'P') dp[0][1] = 1;
    if (a[0] != 'S') dp[0][2] = 1;
    for (int i = 1; i < N; ++i) {
        if (a[i] != 'R') {
            dp[i][0] = (dp[i-1][1] + dp[i-1][2]) % MOD;
        }
        if (a[i] != 'P') {
            dp[i][1] = (dp[i-1][0] + dp[i-1][2]) % MOD;
        }
        if (a[i] != 'S') {
            dp[i][2] = (dp[i-1][0] + dp[i-1][1]) % MOD;
        }
    }
    int ans = (dp[N-1][0] + dp[N-1][1] + dp[N-1][2]) % MOD;
    return ans;
}.

RocK paper scissors ✅