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Lexographically smallest question code in C++ 17 Telegram:- @Infosys_Ans

array AR of size N perfect_sum(arr, s, result) : x = [0]*len(arr) j = len(arr) - 1 while (s > 0) : x[j] = s % 2 s = s // 2 j -= 1 sum = 0 for i in range(len(arr)) : if (x[i] == 1) : sum += arr[i] if (sum == result) : print("{",end=""); for i in range(len(arr)) : if (x[i] == 1) : print(arr[i],end= ", "); print("}, ",end="") def print_subset(arr, K) : x = pow(2, len(arr)) for i in range(1, x): perfect_sum(arr, i, K) # Driver code arr = [ ] n = int(input("Enter length of array : ")) s=int(input("Enter sum : ")) for i in range(n): ele=int(input("Enter element : ")) arr.append(ele) print_subset(arr, s) Python Telegram :-

class CountOccurrences { public static void main(String[] args) { String str = "aabbccd"; char a[]=str.toCharArray(); int cnt = 0; for (int i = 0; i < a.length; i++) { for (int j = i + 1; j < a.length; j++) { if ((a[i] == a[j])) { cnt++; } } } System.out.println(cnt); } } Java Telegram:-

// Infosys // N flowers on a Recatangular pana int ans = 100000000; void solve(vector<int> a, int n, int k, int index, int sum, int maxsum) { if (k == 1) { maxsum = max(maxsum, sum); sum = 0; for (int i = index; i < n; i++) { sum += a[i]; } maxsum = max(maxsum, sum); ans = min(ans, maxsum); return; } sum = 0; for (int i = index; i < n; i++) { sum += a[i]; maxsum = max(maxsum, sum); solve(a, n, k - 1, i + 1, sum, maxsum); } } int GetMaxBeauty(int N, int K, vector<int> A) { solve(A, N, K, 0, 0, 0); return ans; } Telegram:- @allcoding1_official

Number Of Subsequences Code For Infosys int A[] = {10,13,7,8,14,11}; int n = 6; int memo[6];//initialized with -1s; int count(int currIndex){ if (currIndex == n-1) return 1; if (memo[currIndex] != -1) return memo[currIndex]; int res = 0; for (int i=currIndex+1 ; i<n ; i++){ if (abss(A[currIndex] - A[i]) <= 3){ res += count(i); } } memo[currIndex] = res; return res; } c++ Telegram:- @allcoding1_official

Type your code

C++ Telegram -

Wrote correctly...but it shows errors

n = int(input()) A = int(input()) A1 = [] A2= [] A3= [] ans = 0 for i in range(n): A1.append(int(input())) for j in range(n): A2.append(int(input())) for k in range(n): A2.append([p[k], b[k]]) a.sort() for z in a: if a[0] > N: break N += a[1] ans += 1 print(ans) Try this code for this question... Python 3 Telegram:-

Infosys exam Ans 29/7/22 at 10AM http://www.joboffersadda.com/2022/01/infosys-exam-answer.html http://www.joboffersadda.com/2022/01/infosys-exam-answer.html Guys ❤️ Once check it after msg in group Allcoding Ans are available Telegram:- @Infosys_Ans

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