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https://www.allcoding1.com/2023/12/allcoding1-answers.html?m=1

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for_in range(int(input())): n, x = map(int, input().split()) if n == 1 and x == 0: print(1) continue alice = 2 remaining = n -1 if x != 0 and (x == n or x == n - 1): print(-1) elif x == 0: for j in range(1, n + 1): print(j, end=" ") print() else: a = list(range(1, n + 1)) a.pop() a.insert(max(0, n-x-2), n) for j in a: print(j, end="") print() Codechef Greedy Lis

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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 Telegram:- @allcoding1_official

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#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

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class Solution { public: void bfs(vector>&vis,vector>&grid,int i,int j,int n,int m) { vis[i][j]=1; queue>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=0 and ncol>& grid) { int n=grid.size(); int m=grid[0].size(); vector>vis(n,vector(m,0)); int cnt=0; for(int i=0;i

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#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

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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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SELECT category, title, total_stock FROM ( SELECT p.category, p.title, SUM(w.quantity) AS total_stock FROM products p JOIN warehouse w ON p.product_id = w.product_id GROUP BY p.category, p.title HAVING total_stock > 10 ) AS filtered_data ORDER BY category ASC, title ASC, total_stock DESC; IBM✅

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+1

IBM✅ SQL

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+1

return min(e-b+1, numChapters) Python 3✅ IBM Telegram:-

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