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

• Difficulty Level : Basic
• Last Updated : 25 Aug, 2021

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Below is the implementation of above idea.

## C++

 `// C++ program to check if a single bit can``// be flipped tp make all ones``#include ``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`

## PHP

 ``

## Javascript

 ``

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++

 `// 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 ``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`

## PHP

 ``

## Javascript

 ``

Output:

`Yes`

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