en
Feedback
Deloitte | PWC | ATS resumes | Wipro | Infosys | Accenture | Capegemini | job links

Deloitte | PWC | ATS resumes | Wipro | Infosys | Accenture | Capegemini | job links

Open in Telegram
4 673
Subscribers
No data24 hours
-17 days
-630 days
Posts Archive
Hi guys.. 👉🎯Those want help in Wiley edge, Novago , IBM , Accenture nd any placement exams,(or any other ON & OFF campus). COGNIZANT internal exam✅ **Capgemini, Accenture , Hcl, internal exams nd any company internal exams contact us 100 percent selected 🎯 And any certifications(Google, Microsoft, AWS, Pl 300, Pl 400, etc) help contact us 100 percent pass. Link : @PlacementsBro😎 Please refer to ur Frd’s ..  🫶🏻❤️ Channel-- @Coding_Palace

Done airline problem ✅ Fest and posters ✅

Axis bank exam help available Contact @placementsBro😎 100 percent clearance ✅

import java.util.*; public class PrimeSumOptimal {         public static void main(String[] args) {         Scanner sc = new Scanner(System.in);         int N = sc.nextInt();         int[] A = new int[N];         for (int i = 0; i < N; i++) {             A[i] = sc.nextInt();         }         System.out.println(maxNonPrimeSumSubset(A, N));         sc.close();     }     private static boolean[] isPrime;     private static void sieve(int maxLimit) {         isPrime = new boolean[maxLimit + 1];         Arrays.fill(isPrime, true);         isPrime[0] = isPrime[1] = false;         for (int p = 2; p * p <= maxLimit; p++) {             if (isPrime[p]) {                 for (int i = p * p; i <= maxLimit; i += p) {                     isPrime[i] = false;                 }             }         }     }         private static boolean isPrime(int num) {         return isPrime[num];     }     private static int maxNonPrimeSumSubset(int[] A, int N) {         sieve(1000);         int maxSubsetSize = 0;         for (int bitmask = 0; bitmask < (1 << N); bitmask++) {             List<Integer> subset = new ArrayList<>();             for (int i = 0; i < N; i++) {                 if ((bitmask & (1 << i)) != 0) {                     subset.add(A[i]);                 }             }             boolean validSubset = true;             int subsetSize = subset.size();             for (int i = 0; i < subsetSize && validSubset; i++) {                 for (int j = i + 1; j < subsetSize; j++) {                     if (isPrime(subset.get(i) + subset.get(j))) {                         validSubset = false;                         break;                     }                 }             }             if (validSubset) {                 maxSubsetSize = Math.max(maxSubsetSize, subsetSize);             }         }                 return maxSubsetSize;     } }.  pair sum Share @Coding_palace

const int MOD = 1000000007; int subsetSumCount(const vector<int>& A, int L, int R, int K) {     vector<int> dp(K + 1, 0);     dp[0] = 1;     for (int i = L; i <= R; ++i) {         for (int j = K; j >= A[i]; --j) {             dp[j] = (dp[j] + dp[j - A[i]]) % MOD;         }     }     return dp[K]; } int findXOR(int n, int Q, const vector<int>& A, const vector<vector<int>>& B ) {     int result = 0;     for (const auto& query : B ) {         int L = query[0] - 1;         int R = query[1] - 1;         int K = query[2];         int P = subsetSumCount(A, L, R, K);         result ^= P;     }     return result; } //subarray subset sum✅

def is_palindrome(s): return s == s[::-1] def longest_palindrome_from_substrings(A): palindromes = [] pairs = [] max_single_palindrome = "" for s in A: if is_palindrome(s): palindromes.append(s) if len(s) > len(max_single_palindrome): max_single_palindrome = s for i in range(len(A)): for j in range(i + 1, len(A)): combined1 = A[i] + A[j] combined2 = A[j] + A[i] if is_palindrome(combined1): pairs.append(combined1) if is_palindrome(combined2): pairs.append(combined2) longest_palindrome = max_single_palindrome for p in pairs: if len(p) > len(longest_palindrome): longest_palindrome = p return longest_palindrome Palindromic String✅