allcoding1_official

Create rectanglesGiven a rectangle with dimensions N'M and an integer K. You divide this rectangle into smaller sub-rectanglessuch that the given conditions are satisfied: ●Sub-rectangles must be parallel to the axis of the larger rectangle with dimensions N*M.Every sub-rectangle has at least one edge on the larger rectangle edge. Informally, there is no sub-rectangle that is surrounded by other sub- rectangles.For a sub-rectangle with area S, the cost of this sub- rectangle is (S-K)2Calculate the minimum total cost to divide the larger rectangle into smaller sub-rectangles.Note: All sub-rectangles must have an integral length of dimensions.Function description Complete the solve function. This function takes thefollowing 3 parameters and returns the required answer:• N. Represents the value of N • M. Represents the value of M• K: Represents the value of K Input format for custom testingNote: Use this input format if you are testing against custom input or writing code In a language where wedon't provide boilerplate code. . The first line contains 7, which represents thenumber of test cases. . For each test case: o The first line contains an integer N.o The second line contains an integer M. o The third line contains an integer K.Output format For each test case in a new line, print the answerrepresenting the minimum cost. Constraints1≤T≤5 1≤N, M≤ 3001≤K≤ 104 Sample input1 44 15Sample output 1 The first test caseExplanation The first line contains the number of test cases, T = 1.Given • N=4• M = 4 K = 15Approach • You can create one sub-rectangle of dimension 4'4.Thus, the cost of the sub-rectangle is (16-15)^2. . Thus, the minimum cost is 1. The following test cases are the actual test casesof this question that may be used to evaluate your submission. Sample input 1 >1 23 3Sample output 1 0 Sample input 2 1 14 7 Sample output 29 Note:Your code must be able to print the sample output from the provided sample input. However, your code is run against multiple hidden test cases. Therefore, your code must passthese hidden test cases to solve the problem statement. LimitsTime Limit: 3.0 sec(s) for each input file Memory Limit: 256 MBSource Limit: 1024 KB ScoringScore is assigned if any testcase passes Allowed LanguagesGo, Java 8, Java 14, Help me to these code ========================== Answer:- def solve(N, M, K): dp = [[0] * (M + 1) for _ in range(N + 1)] for i in range(1, N + 1): for j in range(1, M + 1): dp[i][j] = float('inf') for k in range(1, i + 1): for l in range(1, j + 1): if k == i and l == j: continue dp[i][j] = min(dp[i][j], (i - k + 1) * (j - l + 1) + dp[k][l] + dp[i - k][j] + dp[k][j - l]) dp[i][j] = min(dp[i][j], K ** 2) return dp[N][M] # Example usage N1, M1, K1 = 4, 4, 15 print(solve(N1, M1, K1)) # Output: 1 N2, M2, K2 = 23, 3, 3 print(solve(N2, M2, K2)) # Output: 0 N3, M3, K3 = 14, 7, 7 print(solve(N3, M3, K3)) # Output: 29

def solve(N, M, K): dp = [[0] * (M + 1) for _ in range(N + 1)] for i in range(1, N + 1): for j in range(1, M + 1): dp[i][j] = float('inf') for k in range(1, i + 1): for l in range(1, j + 1): if k == i and l == j: continue dp[i][j] = min(dp[i][j], (i - k + 1) * (j - l + 1) + dp[k][l] + dp[i - k][j] + dp[k][j - l]) dp[i][j] = min(dp[i][j], K ** 2) return dp[N][M] # Example usage N1, M1, K1 = 4, 4, 15 print(solve(N1, M1, K1)) # Output: 1 N2, M2, K2 = 23, 3, 3 print(solve(N2, M2, K2)) # Output: 0 N3, M3, K3 = 14, 7, 7 print(solve(N3, M3, K3)) # Output: 29

package Graph; import java.util.*; public class Largest_Sum_Cycle { public static int solution(int arr[]) {   ArrayList<Integer>sum=new ArrayList<>();     for(int i=0;i<arr.length;i++)   {       ArrayList<Integer>path=new ArrayList<>();       int j=i;       int t=0;      while(arr[j]<arr.length&&arr[j]!=i&&arr[j]!=-1&&!path.contains(j))   {    path.add(j);    t+=j;    j=arr[j];    if(arr[j]==i)    {     t+=j;     break;    }   }   if(j<arr.length&&i==arr[j])    sum.add(t);   }   if(sum.isEmpty())    return -1;     return Collections.max(sum);   } public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc=new Scanner(System.in); int testcases=sc.nextInt(); for(int loop=0;loop<testcases;loop++) { int numofBlocks=sc.nextInt(); int arr[]=new int[numofBlocks]; int src,dest; for(int i=0;i<numofBlocks;i++) { arr[i]=sc.nextInt(); } System.out.println(solution(arr)); } } } Juspay ✅ Telegram:-

import java.util.Scanner; import java.util.*; public class metting {     public static void helperFunction()     {         Scanner sc = new Scanner(System.in);         int n = sc.nextInt();         int[] edges = new int[n];                 for (int i = 0; i < n; i++)         {             edges[i] = sc.nextInt();         }                 int C1 = sc.nextInt();         int C2 = sc.nextInt();         // int ans=minimumWeight(n,edges,C1,C2);         // System.out.println(ans);   //  public static int minimumWeight(int n, int[] edges, int C1, int C2) {         List<List<Integer>> list = new ArrayList<>();         for (int i = 0; i < n; i++) {             list.add(new ArrayList<Integer>());         }         for (int i = 0; i < n; i++) {             if (edges[i] != -1) {                 list.get(i).add(edges[i]);             }         }         long[] array1 = new long[n];         long[] array2 = new long[n];         Arrays.fill(array1, Long.MAX_VALUE);         Arrays.fill(array2, Long.MAX_VALUE);         juspay(C1, list, array1);         juspay(C2, list, array2);         int node = 0;         long dist = Long.MAX_VALUE;         for (int i = 0; i < n; i++) {             if (array1[i] == Long.MAX_VALUE || array2[i] == Long.MAX_VALUE)                 continue;             if (dist > array1[i] + array2[i]) {                 dist = array1[i] + array2[i];                 node = i;             }         }         if (dist == Long.MAX_VALUE)         System.out.print(-1);             //return -1;        // return node;          System.out.print(node);     }     private static void juspay(int start, List<List<Integer>> graph, long[] distances)     {         PriorityQueue<Integer> pq = new PriorityQueue<>();         pq.offer(start);         distances[start] = 0;         while (!pq.isEmpty())         {             int curr = pq.poll();             for (int neighbor : graph.get(curr))             {                 long distance = distances[curr] + 1;                 if (distance < distances[neighbor])                 {                     distances[neighbor] = distance;                     pq.offer(neighbor);                 }             }         }     }     public static void main(String[] args) {      metting m = new metting();         metting.helperFunction();     } } Nearest meeting Cell Juspay ✅

🎯TCS National Qualifier Test (TCS NQT) 2024 Location: Across India Qualification: B.E / B.Tech / M.E / M.Tech / M.Sc / MCA / Any Graduate / Under Graduate / Diploma Batch: 2018/2019/2020/2021/2022/2023/2024 Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official

Copy and paste question

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

any paid promotion DM :- @Priya_i

Fake #scammer @allcoding1_officiaI

KPMG Off Campus Drive Hiring 2024 | Front End Developer | 4 LPA+ Job Title : Front End Developer Qualification : B.E / B.Tech Batch : Any Batch Package : 4 LPA+ Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official

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✅

IBM✅ SQL

🎯Google Internship 2024 | Software Student Training in Engineering Program | Apply Now Job Role : Software Student Training in Engineering Program Qualification : B.E/B.Tech/B.Sc Experience : Freshers Last Date : 19 January 2024 Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official

🎯Phone pe hiring Job role:- Advisor,ONDC Qualification:- Any Experience:- 0-2years Location:- Bangalore Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official