Given a binary string str, the task is to determine if string str can be converted into a palindrome in K moves. In one move any one bit can be flipped i.e. 0 to 1 or 1 to 0.
Examples:
Input: str = “101100”, K = 1
Output: YES
Explanation: Flip last bit of str from 0 to 1.Input: str = “0101101”, K = 2
Output: NO
Explanation: Three moves required to make str a palindrome.
Approach: The idea is to traverse the string with two pointers.
- Match the bits pointed by both pointers and
- Keep count of the failed comparisons.
- Number of unmatched pairs will be the desired output.
Below is the implementation of the above approach:
C++
// C++ program to check// if given binary string// can be made palindrome in K moves#include <bits/stdc++.h>using namespace std;// Function to make Binary string// palindrome in K stepsint MinFlipPalindrome(string str, int K){ int n = str.size(); int i = 0, j = n - 1, count = 0; while (i <= j) { if (str[i] == str[j]) { i += 1; j -= 1; continue; } else { i += 1; j -= 1; count += 1; } } if (count <= K) return 1; else return 0;}// Driver codeint main(){ string str = "101011110"; int K = 2; int res = MinFlipPalindrome(str, K); if (res == 1) cout << "YES"; else cout << "NO"; return 0;} |
C
// C program to check if given binary string// can be made palindrome in K moves#include <stdio.h>#include <string.h>// Function to make Binary string// palindrome in K stepsint MinFlipPalindrome(char* str, int K){ int n = strlen(str); int i = 0, j = n - 1, count = 0; while (i <= j) { if (str[i] == str[j]) { i += 1; j -= 1; continue; } else { i += 1; j -= 1; count += 1; } } if (count <= K) return 1; else return 0;}// Driver codeint main(){ char* str = "101011110"; int K = 2; int res = MinFlipPalindrome(str, K); printf("%s\n", (res ? "YES" : "NO")); return 0;}// This code is contributed by phalasi. |
Java
// Java program to check// if given binary string// can be made palindrome in K movesimport java.io.*;class GFG { // Function to make Binary string // palindrome in K steps static int MinFlipPalindrome(String str, int K) { int n = str.length(); int i = 0, j = n - 1, count = 0; while (i <= j) { if (str.charAt(i) == str.charAt(j)) { i += 1; j -= 1; continue; } else { i += 1; j -= 1; count += 1; } } if (count <= K) return 1; else return 0; } // Driver code public static void main(String[] args) { String str = "101011110"; int K = 2; int res = MinFlipPalindrome(str, K); if (res == 1) System.out.println("YES"); else System.out.println("NO"); }}// This code is contributed by Karandeep Singh |
Python3
# Python program for the above approach# Function to make Binary string# palindrome in K stepsdef MinFlipPalindrome(str, K): n = len(str) i = 0 j = n - 1 count = 0 while (i <= j): if (str[i] == str[j]): i += 1 j -= 1 continue else: i += 1 j -= 1 count += 1 if (count <= K): return 1 else: return 0# Driver codestr = "101011110"K = 2res = MinFlipPalindrome(str, K)if (res == 1): print("YES")else: print("NO") # This code is contributed by sanjoy_62. |
C#
// C# program to check// if given binary string// can be made palindrome in K movesusing System;class GFG { // Function to make Binary string // palindrome in K steps static int MinFlipPalindrome(string str, int K) { int n = str.Length; int i = 0, j = n - 1, count = 0; while (i <= j) { if (str[i] == str[j]) { i += 1; j -= 1; continue; } else { i += 1; j -= 1; count += 1; } } if (count <= K) return 1; else return 0; } // Driver code public static void Main() { string str = "101011110"; int K = 2; int res = MinFlipPalindrome(str, K); if (res == 1) Console.Write("YES"); else Console.Write("NO"); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript program to check // if given binary string // can be made palindrome in K moves // Function to make Binary string // palindrome in K steps const MinFlipPalindrome = (str, K) => { let n = str.length; let i = 0, j = n - 1, count = 0; while (i <= j) { if (str[i] == str[j]) { i += 1; j -= 1; continue; } else { i += 1; j -= 1; count += 1; } } if (count <= K) return 1; else return 0; } // Driver code let str = "101011110"; let K = 2; let res = MinFlipPalindrome(str, K); if (res == 1) document.write("YES"); else document.write("NO"); // This code is contributed by rakeshsahni </script> |
Output
NO
Time complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
