Check if all bits can be made same by single flip

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Given a binary string, find if it is possible to make all its digits equal (either all 0’s or all 1’s) by flipping exactly one bit.

Input: 101
Output: Yes
Explanation: In 101, the 0 can be flipped
             to make it all 1
Input: 11
Output: No
Explanation: No matter whichever digit you 
  flip, you will not get the desired string.
Input: 1
Output: Yes
Explanation: We can flip 1, to make all 0's

Method 1 (Counting 0’s and 1’s)
If all digits of a string can be made identical by doing exactly one flip, that means the string has all its digits equal to one another except this digit which has to be flipped, and this digit must be different than all other digits of the string. The value of this digit could be either zero or one. Hence, this string will either have exactly one digit equal to zero, and all other digits equal to one, or exactly one digit equal to one, and all other digit equal to zero.
Therefore, we only need to check whether the string has exactly one digit equal to zero/one, and if so, the answer is yes; otherwise the answer is no.

Below is the implementation of above idea.

C++

// C++ program to check if a single bit can
// be flipped tp make all ones
#include <bits/stdc++.h>
using namespace std;
 
// This function returns true if we can
// bits same in given binary string str.
bool canMakeAllSame(string str)
{
    int zeros = 0, ones = 0;
 
    // Traverse through given string and
    // count numbers of 0's and 1's
    for (char ch : str)
        (ch == '0') ? ++zeros : ++ones;
 
    // Return true if any of the two counts
    // is 1
    return (zeros == 1 || ones == 1);
}
 
// Driver code
int main()
{
    canMakeAllSame("101") ? printf("Yes\n") : printf("No\n");
    return 0;
}

Java

// Java program to check if a single bit can
// be flipped to make all ones
public class GFG {
 
    // This function returns true if we can
    // bits same in given binary string str.
    static boolean canMakeAllSame(String str)
    {
        int zeros = 0, ones = 0;
 
        // Traverse through given string and
        // count numbers of 0's and 1's
        for (int i = 0; i < str.length(); i++) {
            char ch = str.charAt(i);
            if (ch == '0')
                ++zeros;
            else
                ++ones;
        }
 
        // Return true if any of the two counts
        // is 1
        return (zeros == 1 || ones == 1);
    }
 
    // Driver code
    public static void main(String args[])
    {
        System.out.println(canMakeAllSame("101") ? "Yes" : "No");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# python program to check if a single
# bit can be flipped tp make all ones
 
# This function returns true if we can
# bits same in given binary string str.
def canMakeAllSame(str):
    zeros = 0
    ones = 0
 
    # Traverse through given string and
    # count numbers of 0's and 1's
    for i in range(0, len(str)):
        ch = str[i];
        if (ch == '0'):
            zeros = zeros + 1
        else:
            ones = ones + 1
 
    # Return true if any of the two
    # counts is 1
    return (zeros == 1 or ones == 1);
 
# Driver code
if(canMakeAllSame("101")):
    print("Yes\n")
else:
    print("No\n")
 
# This code is contributed by Sam007.

C#

// C# program to check if a single bit can
// be flipped to make all ones
using System;
 
class GFG {
     
    // This function returns true if we can
    // bits same in given binary string str.
    static bool canMakeAllSame(string str)
    {
        int zeros = 0, ones = 0;
 
        // Traverse through given string and
        // count numbers of 0's and 1's
        for (int i = 0; i < str.Length; i++) {
            char ch = str[i];
            if (ch == '0')
                ++zeros;
            else
                ++ones;
        }
 
        // Return true if any of the two counts
        // is 1
        return (zeros == 1 || ones == 1);
    }
 
    // Driver code
    public static void Main()
    {
        Console.WriteLine(canMakeAllSame("101") ? "Yes" : "No");
    }
}
 
// This code is contributed by Sam007

Javascript

<script>
// Javascript program to check if a single bit can
// be flipped to make all ones
     
    // This function returns true if we can
    // bits same in given binary string str.
    function canMakeAllSame(str)
    {
        let zeros = 0, ones = 0;
   
        // Traverse through given string and
        // count numbers of 0's and 1's
        for (let i = 0; i < str.length; i++) {
            let ch = str[i];
            if (ch == '0')
                ++zeros;
            else
                ++ones;
        }
   
        // Return true if any of the two counts
        // is 1
        return (zeros == 1 || ones == 1);
    }
     
    // Driver code
    document.write(canMakeAllSame("101") ? "Yes" : "No");
     
    // This code is contributed by rag2127
</script>

PHP

<?php
 
// PHP program to check if a single bit can
// be flipped tp make all ones
  
// This function returns true if we can
// bits same in given binary string str.
function canMakeAllSame($str)
{
    $zeros = 0;
    $ones = 0;
  
    // Traverse through given string and
    // count numbers of 0's and 1's
     
    for($i=0;$i<strlen($str);$i++)
    {
        $ch = $str[$i];
        if ($ch == '0') 
            ++$zeros ;
        else
            ++$ones;
    }
    // Return true if any of the two counts
    // is 1
    return ($zeros == 1 || $ones == 1);
}
  
// Driver code
 
    if (canMakeAllSame("101") )
       echo "Yes\n" ;
    else echo "No\n";
    return 0;
?>

Output:

Yes

Time complexity : O(n) where n is the length of the string.

Auxiliary Space: O(1)

Method 2 (Counting 0’s and 1’s)
The idea is to compute sum of all bits. If sum is n-1 or 1, then output is true, else false. This solution doesn’t require a comparison in a loop.

Below is the implementation of above idea.

C++

// Check if all bits can be made same by single flip
// Idea is to add the integer value all the elements
// in the given string.
// If the sum is 1 it indicates that there is
// only single '1' in the string.
// If the sum is 0 it indicates that there is only
// single '0' in the string.
// It takes O(n) time.
 
#include <bits/stdc++.h>
using namespace std;
 
bool isOneFlip(string str)
{
    int sum = 0;
    int n = str.length();
 
    // Traverse through given string and
    // count the total sum of numbers
    for (int i = 0; i < n; i++)
        sum += str[i] - '0';
 
    // Return true if any of the two counts
    // is 1
    return (sum == n - 1 || sum == 1);
}
 
// Main function
int main()
{
    isOneFlip("101111111111") ? printf("Yes\n") : printf("No\n");
 
    return 0;
}

Java

/*Check if all bits can be made same by single 
flip. Idea is to add the integer value all the
elements in the given string.
If the sum is 1 it indicates that there is
   only single '1' in the string.
If the sum is 0 it indicates that there is only
   single '0' in the string.
It takes O(n) time.*/
public class GFG {
 
    static boolean isOneFlip(String str)
    {
        int sum = 0;
        int n = str.length();
 
        // Traverse through given string and
        // count the total sum of numbers
        for (int i = 0; i < n; i++)
            sum += str.charAt(i) - '0';
 
        // Return true if any of the two counts
        // is 1
        return (sum == n - 1 || sum == 1);
    }
 
    // Main function
    public static void main(String args[])
    {
        System.out.println(isOneFlip("101111111111") ? "Yes" : "No");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Check if all bits can be made same
# by single flip Idea is to add the 
# integer value all the elements in
# the given string. If the sum is 1
# it indicates that there is only
# single '1' in the string. If the
# sum is 0 it indicates that there
# is only single '0' in the string.
# It takes O(n) time.
 
def isOneFlip(str):
 
    sum = 0
    n = len(str)
 
    # Traverse through given string
    # and count the total sum of
    # numbers
    for i in range( 0, n ):
        sum += int(str[i]) - int('0')
 
    # Return true if any of the two
    # counts is 1
    return (sum == n - 1 or sum == 1)
 
# Main function
(print("Yes") if isOneFlip("101111111111") 
                        else print("No"))
 
# This code is contributed by Smitha

C#

/*Check if all bits can be made same by single 
  flip. Idea is to add the integer value all the
  elements in the given string.
  If the sum is 1 it indicates that there is
  only single '1' in the string.
  If the sum is 0 it indicates that there is only
  single '0' in the string.
  It takes O(n) time.*/
using System;
 
class GFG {
     
    static bool isOneFlip(string str)
    {
        int sum = 0;
        int n = str.Length;
 
        // Traverse through given string and
        // count the total sum of numbers
        for (int i = 0; i < n; i++)
            sum += str[i] - '0';
 
        // Return true if any of the two counts
        // is 1
        return (sum == n - 1 || sum == 1);
    }
 
    // Driver code
    public static void Main()
    {
        Console.WriteLine(isOneFlip("101111111111") ? "Yes" : "No");
    }
}
 
// This code is contributed by Sam007

Javascript

<script>
 
/*Check if all bits can be made same by single
flip. Idea is to add the integer value all the
elements in the given string.
If the sum is 1 it indicates that there is
   only single '1' in the string.
If the sum is 0 it indicates that there is only
   single '0' in the string.
It takes O(n) time.*/
function isOneFlip(str)
{
    let sum = 0;
    let n = str.length;
 
    // Traverse through given string and
    // count the total sum of numbers
    for(let i = 0; i < n; i++)
        sum += str[i] - '0';
 
    // Return true if any of the two counts
    // is 1
    return (sum == n - 1 || sum == 1);
}
 
// Driver code
document.write(isOneFlip("101111111111") ?
               "Yes" : "No");
 
// This code is contributed by avanitrachhadiya2155
 
</script>

PHP

<?php
// Check if all bits can be made same by
// single flip Idea is to add the integer
// value all the elements in the given 
// string. If the sum is 1 it indicates
// that there is only single '1' in the
// string. If the sum is 0 it indicates
// that there is only single '0' in the
// string. It takes O(n) time.
 
function isOneFlip($str)
{
    $sum = 0;
    $n = strlen($str);
 
    // Traverse through given string and
    // count the total sum of numbers
    for ( $i = 0; $i < $n; $i++)
        $sum += $str[$i] - '0';
 
    // Return true if any of the two counts
    // is 1
    return ($sum == $n - 1 || $sum == 1);
}
 
// Main function
    if(isOneFlip("101111111111") )
        echo "Yes\n";
    else
        echo "No\n";
 
// This code is contributed by aj_36
?>

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)
Thanks to Sourabh Gavhale for suggesting this solution

Method 3 (Iterating through string)
The idea is to iterate through the string and for each position, check if flipping that bit would make all the bits in the string the same. If yes, then return true, else continue iterating. If no such position is found, return false.

Below is the implementation of the above idea.

C++

// Check if all bits can be made same by single flip
// Idea is to iterate through each bit in the string
// and check if flipping that bit would make all the
// bits same. It takes O(n) time.
 
#include <bits/stdc++.h>
using namespace std;
 
bool isOneFlip(string str)
{
    int n = str.length();
 
// Iterate through each bit in the string
    for (int i = 0; i < n; i++) {
    // Flip the current bit
        str[i] = (str[i] == '0') ? '1' : '0';
 
        bool allSame = true;
        char firstBit = str[0];
 
    // Check if all bits in the string are the same
        for (int j = 1; j < n; j++) {
            if (str[j] != firstBit) {
                allSame = false;
                break;
            }
        }
 
    // If all bits are the same, return true
    if (allSame) {
        return true;
    }
 
    // Flip the bit back to its original value
    str[i] = (str[i] == '0') ? '1' : '0';
}
 
// No single flip can make all bits same
return false;
}
 
// Main function
int main()
{
    isOneFlip("101") ? printf("Yes\n") : printf("No\n");
    return 0;
}

Java

import java.util.*;
 
public class Main {
 
    public static boolean isOneFlip(String str) {
        int n = str.length();
 
        // Iterate through each bit in the string
        for (int i = 0; i < n; i++) {
            // Flip the current bit
            char currentBit = (str.charAt(i) == '0') ? '1' : '0';
            StringBuilder modifiedStr = new StringBuilder(str);
            modifiedStr.setCharAt(i, currentBit);
 
            boolean allSame = true;
            char firstBit = modifiedStr.charAt(0);
 
            // Check if all bits in the modified string are the same
            for (int j = 1; j < n; j++) {
                if (modifiedStr.charAt(j) != firstBit) {
                    allSame = false;
                    break;
                }
            }
 
            // If all bits are the same, return true
            if (allSame) {
                return true;
            }
        }
 
        // No single flip can make all bits the same
        return false;
    }
 
    // Main function
    public static void main(String[] args) {
        boolean result = isOneFlip("101");
        System.out.println(result ? "Yes" : "No");
    }
}

Python3

def is_one_flip(s):
    n = len(s)
 
    # Iterate through each bit in the string
    for i in range(n):
        # Flip the current bit
        s = s[:i] + ('0' if s[i] == '1' else '1') + s[i + 1:]
 
        all_same = True
        first_bit = s[0]
 
        # Check if all bits in the string are the same
        for j in range(1, n):
            if s[j] != first_bit:
                all_same = False
                break
 
        # If all bits are the same, return True
        if all_same:
            return True
 
        # Flip the bit back to its original value
        s = s[:i] + ('0' if s[i] == '1' else '1') + s[i + 1:]
 
    # No single flip can make all bits the same
    return False
 
# Main function
if is_one_flip("101"):
    print("Yes")
else:
    print("No")
 
    # This code is contributed by shivamgupta310570

C#

// Check if all bits can be made same by single flip
// Idea is to iterate through each bit in the string
// and check if flipping that bit would make all the
// bits same. It takes O(n) time.
using System;
 
class Program
{
    static bool IsOneFlip(string str)
    {
        int n = str.Length;
 
          // Iterate through each bit in the string
        for (int i = 0; i < n; i++)
        {
              // Flip the current bit
            str = str.Substring(0, i) + (str[i] == '0' ? '1' : '0') + str.Substring(i + 1);
            bool allSame = true;
            char firstBit = str[0];
 
              // Check if all bits in the string are the same
            for (int j = 1; j < n; j++)
            {
                if (str[j] != firstBit)
                {
                    allSame = false;
                    break;
                }
            }
 
              // If all bits are the same, return true
            if (allSame)
            {
                return true;
            }
 
              // Flip the bit back to its original value
            str = str.Substring(0, i) + (str[i] == '0' ? '1' : '0') + str.Substring(i + 1);
        }
 
          // No single flip can make all bits same
        return false;
    }
 
      // Main function
    static void Main()
    {
        Console.WriteLine(IsOneFlip("101") ? "Yes" : "No");
    }
}

Output

Yes

Time Complexity: O(N^2) in worst case
Auxiliary Space: O(1)

Method 4 (XOR approach)
The idea is to compute the XOR of all the bits in the string. If the XOR is either 0 or 1, then return true, else return false.

Below is the implementation of the above idea.

C++

#include <bits/stdc++.h>
using namespace std;
 
bool isOneFlip(string str)
{
    int n = str.length();
    int xorVal = 0;
 
// Compute XOR of all bits in the string
    for (int i = 0; i < n; i++) {
        xorVal ^= (str[i] - '0');
    }
 
// If XOR is 0 or 1, return true
    return (xorVal == 0 || xorVal == 1);
}
 
// Main function
int main()
{
    isOneFlip("101") ? printf("Yes\n") : printf("No\n");
    return 0;
}

Java

import java.io.*;
 
 
public class GFG {
 
    static boolean isOneFlip(String str) {
        int n = str.length();
        int xorVal = 0;
 
        // Compute XOR of all bits in the string
        for (int i = 0; i < n; i++) {
            xorVal ^= (str.charAt(i) - '0');
        }
 
        // If XOR is 0 or 1, return true
        return (xorVal == 0 || xorVal == 1);
    }
 
    // Main function
    public static void main(String[] args) {
        if (isOneFlip("101")) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

This article is contributed by Aarti_Rathi and Subrata Ghosh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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