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import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class RequestParser {
    
    public static List<string> getResponses(String[] validAuthTokens, String[][] requests) {
        List<string> responses = new ArrayList&lt;&gt;();

        Set<string> validTokensSet = new HashSet&lt;&gt;(Arrays.asList(validAuthTokens));
        
        for (String[] request : requests) {
            String requestType = request[0];
            String url = request[1];
            
            // Extract the token and other parameters from the URL
            String[] urlParts = url.split("\\?");
            String[] params = urlParts[1].split("&amp;");
            
            String token = null;
            String csrf = null;
            StringBuilder otherParams = new StringBuilder();
            
            for (String param : params) {
                String[] keyValue = param.split("=");
                if (keyValue[0].equals("token")) {
                    token = keyValue[1];
                } else if (keyValue[0].equals("csrf")) {
                    csrf = keyValue[1];
                } else {
                    otherParams.append(",").append(keyValue[0]).append(",").append(keyValue[1]);
                }
            }
            
            // Validate the authentication token
            if (!validTokensSet.contains(token)) {
                responses.add("INVALID");
                continue;
            }
            
            // Validate the request type and CSRF token for POST requests
            if (requestType.equals("POST")) {
                if (csrf == null || !csrf.matches("[a-z0-9]{8,}")) {
                    responses.add("INVALID");
                    continue;
                }
            }
            
            // If all validations pass, construct the response string
            if (otherParams.length() &gt; 0) {
                responses.add("VALID" + otherParams);
            } else {
                responses.add("VALID");
            }
        }
        
        return responses;
    }

    public static void main(String[] args) {
        String[] validAuthTokens = {"ah37j2ha483u", "safh34ywb0p5", "ba34wyi8t902"};
        String[][] requests = {
            {"GET", "https://example.com/?token=347sd6yk8iu2&amp;name=alex"},
            {"GET", "https://example.com/?token=safh34ywb0p5&amp;name=sam"},
            {"POST", "https://example.com/?token=safh34ywb0p5&amp;name=alex"},
            {"POST", "https://example.com/?token=safh34ywb0p5&amp;csrf=ak2sh32dy&amp;name=chri$"}
        };

        List<string> responses = getResponses(validAuthTokens, requests);
        System.out.println(responses);  // Output: [INVALID, VALID,name,sam, INVALID, VALID,name,chri$]
    }
}

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Saks subarray product✅ long long solve(vector& nums, int k) {     if (k <= 1) return 0;     int n = nums.size();     long long p = 1;     int i = 0, j = 0;     long long ans = 0;     while (j < n) {         p *= nums[j];         while (i <= j && p > k) {             p /= nums[i];             i++;         }         ans += j - i + 1;         j++;     }     return ans; }

#include<bits/stdc++.h> using namespace std; vector<int> solve(int N, vector<int> a, int q, vector<vector<int>> queries) {     vector<int> res;     for(int i=0; i<q; i++) {         int T = queries[i][0];         int l = queries[i][1];         int r = queries[i][2];         int x = queries[i][3];         if(T == 1) {             int ans = a[l-1];             for(int j=l; j<r; j++) {                 ans |= a[j];             }             res.push_back(ans);         } else {             for(int j=l-1; j<r; j++) {                 a[j] ^= x;             }         }     }     return res; } int main() {     ios_base::sync_with_stdio(false);     cin.tie(NULL);     int N;     cin>>N;     vector<int> a(N);     for(int i=0; i<N; i++) {         cin>>a[i];     }     int q;     cin>>q;     vector<vector<int>> queries(q, vector<int>(4));     for(int i=0; i<q; i++) {         for(int j=0; j<4; j++) {             cin>>queries[i][j];         }     }     vector<int> res = solve(N, a, q, queries);     for(int i=0; i<res.size(); i++) {         cout<<res[i]<<" ";     }     cout<<endl;     return 0; } Array bitwise operations✅

#include<bits/stdc++.h> using namespace std; class hm {     unordered_map<int, int> m;     int k = 0, v = 0; public:     void i(int x, int y) {         x -= k;         y -= v;         m[x] = y;     }     int g(int x) {         x -= k;         if (m.find(x) == m.end()) {             return -1;         }         return m[x] + v;     }     void ak(int x) {         k += x;     }     void av(int y) {         v += y;     } }; int solve(vector<string>& qt, vector<vector<int>>& q) {     hm h;     int s = 0;     for (int i = 0; i < qt.size(); ++i) {         if (qt[i] == "insert") {             h.i(q[i][0], q[i][1]);         } else if (qt[i] == "addToKey") {             h.ak(q[i][0]);         } else if (qt[i] == "addToValue") {             h.av(q[i][0]);         } else if (qt[i] == "get") {             int v = h.g(q[i][0]);             if (v != -1) {                 s += v;             }         }     }     return s; } Telegram:-

#include <iostream> #include <vector> #include <unordered_map> class Main { public:     static long getZeroBitSubarrays(const std::vector<int>& arr) {         int n = arr.size();         long totalSubarrayCount = static_cast<long>(n) * (n + 1) / 2;         long nonzeroSubarrayCount = 0;         std::unordered_map<int, int> windowBitCounts;         int leftIdx = 0;         for (int rightIdx = 0; rightIdx < n; rightIdx++) {             int rightElement = arr[rightIdx];             if (rightElement == 0) {                 windowBitCounts.clear();                 leftIdx = rightIdx + 1;                 continue;             }             std::vector<int> setBitIndices = getSetBitIndices(rightElement);             for (int index : setBitIndices) {                 windowBitCounts[index]++;             }             while (leftIdx < rightIdx && isBitwiseAndZero(rightIdx - leftIdx + 1, windowBitCounts)) {                 for (int index : getSetBitIndices(arr[leftIdx])) {                     windowBitCounts[index]--;                     if (windowBitCounts[index] == 0) {                         windowBitCounts.erase(index);                     }                 }                 leftIdx++;             }             nonzeroSubarrayCount += (rightIdx - leftIdx + 1);         }         return totalSubarrayCount - nonzeroSubarrayCount;     } private:     static std::vector<int> getSetBitIndices(int x) {         std::vector<int> setBits;         int pow2 = 1;         int exponent = 0;         while (pow2 <= x) {             if ((pow2 & x) != 0) {                 setBits.push_back(exponent);             }             exponent++;             pow2 *= 2;         }         return setBits;     }     static bool isBitwiseAndZero(int windowLength, const std::unordered_map<int, int>& bitCounts) {         for (const auto& entry : bitCounts) {             if (entry.second >= windowLength) {                 return false;             }         }         return true;     } }; DE Shaw ✅ C++

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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 ✅

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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