Check if given Binary String can be made Palindrome using K flips

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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 steps
int 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 code
int 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 steps
int 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 code
int 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 moves
import 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 steps
 
 
def 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 code
str = "101011110"
K = 2
res = 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 moves
using 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)

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