# Change if all bits can be made same by single flip

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 sinle 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
```

Output:

```Yes
```

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

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++

```//Change 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

```/*Change 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
```

Output:

```Yes
```

Thanks to Sourabh Gavhale for suggesting this solution

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

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