def subsetA(arr):
    arr.sort(reverse=True)
    subset_a = []

    sum_a = 0
    sum_b = 0

    for num in arr:
        if sum_a <= sum_b:
            subset_a.append(num)
            sum_a += num
        else:
            sum_b += num

    return sorted(subset_a)

Python 3✅
Array subset

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class Solution {
public:
     void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)
    {
        vis[i][j]=1;
        queue<pair<int,int>>q;
        q.push({i,j});
        while(!q.empty())
        {
            int row=q.front().first;
            int col=q.front().second;
            q.pop();
            int delrow[4]={1,0,-1,0};
            int delcol[4]={0,1,0,-1};
                   for(int k=0;k<=3;k++){
                    int nrow=row+delrow[k];
                    int ncol=col+delcol[k];
                    if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])
                    {
                        vis[nrow][ncol]=1;
                        q.push({nrow,ncol});
                    }
                }
            }
        }
    int numIslands(vector<vector<char>>& grid) {
    int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>vis(n,vector<int>(m,0));
        int cnt=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(!vis[i][j] and grid[i][j]=='1')
                {
                    cnt++;
                    bfs(vis,grid,i,j,n,m);
                }
            }
        }
        return cnt;
    }
};

Zeta ✅

#include <bits/stdc++.h>
using namespace std;

vector<int> solution(vector<int> a, int n, int k) {
    vector<int> v;
    deque<int> dq;

    for (int i = 0; i < n; i++) {
        while (!dq.empty() && dq.front() <= i - k)
            dq.pop_front();

        while (!dq.empty() && a[dq.back()] <= a[i])
            dq.pop_back();

        dq.push_back(i);

        if (i >= k - 1)
            v.push_back(a[dq.front()]);
    }

    return v;
}

int main() {
    int n, k;
    cin >> n >> k;

    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];

    vector<int> result = solution(a, n, k);

    for (int i = 0; i < result.size(); i++)
        cout << result[i] << " ";

    return 0;
}. 

//cricket match ✅
Zeta

string make_string_S_to_T(string S) {
    string T=“programming”;
    bool possible = false;
    int M = T.length();
    int N = S.length();
    for (int i = 0; i <= M; i++) {
        int prefix_length = i;
        int suffix_length = M - i;
        string prefix = S.substr(0, prefix_length);
        string suffix = S.substr(N - suffix_length, suffix_length);
        if (prefix + suffix == T) {
            possible = true;
            break;
        }
    }
    if (possible)
        return "YES";
    else
        return "NO";
}

Deleting substring ✅
Zeta

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