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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

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

#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

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

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

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