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# Mixing proteins – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

# Solutions of Algorithms Data Structures Hard HackerRank:

## Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

# C++ replace HackerRank Solution

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <algorithm> #include <vector> using namespace std; #define NMAX 1111111 char ch; int A[2][NMAX]; int N, K, i, b, p, c, step; int main() { // freopen("x.txt", "r", stdin); scanf("%d %d", &N, &K); for (i = 0; i < N; i++) { scanf(" %c", &ch); A[0][i] = ch - 'A'; } for (c = 0, b = 30; b >= 0; b--) if (K & (1 << b)) { p = c; c = 1 - c; step = (1 << b) % N; for (i = 0; i < N; i++) A[c][i] = A[p][i] ^ A[p][(i + step) % N]; } for (i = 0; i < N; i++) printf("%c", A[c][i] + 'A'); printf("\n"); return 0; }

# Java rep HackerRank Solution

import java.io.*; public class Solution { public static void solve(Input in, PrintWriter out) throws IOException { int n = in.nextInt(), k = in.nextInt(); int[] a = new int[n]; char[] c = in.next().toCharArray(); for (int i = 0; i < n; ++i) { a[i] = c[i] - 'A'; } for (int i = 1; i <= k; i *= 2) { if ((k & i) != 0) { int[] b = new int[n]; for (int j = 0; j < n; ++j) { b[j] = a[j] ^ a[(j + i) % n]; } a = b; } } for (int i = 0; i < n; ++i) { c[i] = (char) (a[i] + 'A'); } out.println(c); } public static void main(String[] args) throws IOException { PrintWriter out = new PrintWriter(System.out); solve(new Input(new BufferedReader(new InputStreamReader(System.in))), out); out.close(); } static class Input { BufferedReader in; StringBuilder sb = new StringBuilder(); public Input(BufferedReader in) { this.in = in; } public Input(String s) { this.in = new BufferedReader(new StringReader(s)); } public String next() throws IOException { sb.setLength(0); while (true) { int c = in.read(); if (c == -1) { return null; } if (" \n\r\t".indexOf(c) == -1) { sb.append((char)c); break; } } while (true) { int c = in.read(); if (c == -1 || " \n\r\t".indexOf(c) != -1) { break; } sb.append((char)c); } return sb.toString(); } public int nextInt() throws IOException { return Integer.parseInt(next()); } public long nextLong() throws IOException { return Long.parseLong(next()); } public double nextDouble() throws IOException { return Double.parseDouble(next()); } } }

# Python 2 rep HackerRank Solution

# A = 0 (0b00) # B = 1 (0b01) # C = 2 (0b10) # D = 3 (0b11) # a[x + 2^m, i] = a[x, i] ^ a[x, i + 2^m] import sys c2i = {'A': 0, 'B': 1, 'C': 2, 'D': 3} i2c = ['A', 'B', 'C', 'D'] n, k = [int(x) for x in input().strip().split()] s = [c2i[x] for x in input().strip()] s = [s, [0] * len(s)] a, b = 1, 0 while k > 0: # swap a ^= b b ^= a a ^= b # closest power of 2 cp2 = 1 for i in range(29, 0, -1): if k - (1 << i) >= 0: cp2 = 1 << i break k -= cp2 for i in range(n): s[b][i] = s[a][i] ^ s[a][(i + cp2) % n] for i in s[b]: sys.stdout.write(i2c[i]) sys.stdout.flush()

# Python 3 rep HackerRank Solution

Str = "ABCD" N,M = [int(x) for x in raw_input().split()] A = raw_input() Len = len(A) B = [ord(A[i])-ord("A") for i in xrange(Len)] C = [0 for i in xrange(Len)] while M!=0: G = M&-M M^=G for i in xrange(Len): C[i]=B[i]^B[(i+G)%Len] for i in xrange(Len): B[i]=C[i] C[i]=0 A = "" for i in xrange(Len): A += Str[B[i]] print(A)

# C rep HackerRank Solution

#include <stdio.h> #include <stdlib.h> int main() { int i, j, n, k, c, *prot, *protnew, p2; scanf("%d %d",&n,&k); prot = malloc(n * sizeof(int)); protnew = malloc(n * sizeof(int)); for (i=0; i<n; i++) { while (c=getchar()-'A') if (c>=0 && c<=3) break; prot[i] = c; } p2 = 1; while (p2<k) p2 <<= 1; while (p2 >>= 1) { if (p2 & k) { for (i=0; i<n; i++) protnew[i] = prot[i] ^ prot[(i+p2) % n]; for (i=0; i<n; i++) prot[i] = protnew[i]; } } for (i=0; i<n; i++) putchar('A' + prot[i]); putchar('\n'); return 0; }

## Warmup

Implementation

Strings

Sorting

Search

Graph Theory

Greedy

Dynamic Programming

Constructive Algorithms

Bit Manipulation

Recursion

Game Theory

NP Complete

Debugging

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