# Check if all bits can be made same by flipping two consecutive bits

Given a binary string, the task is to find whether all the digits of the string can be made equal i.e either 0 or 1 by flipping two consecutive bits any number of times.**Examples:**

Input:01011Output:YESExplanation:Flip 2nd and 3rd bit -> 00111, again flipping 1'st and 2'nd bit -> 11111Input:100011Output:NOExplanation:No number of moves can ever equalize all elements of the array.

**Approach:**

On careful observation, toggling of i’th and j’th bit can be done by toggling from i’th bit like (i, i+1), (i+1, i+2) …. (j-1, j) here every bit is toggling twice (if bit is toggle twice then its come to its initial value) except i and j then ultimately i’th and j’th bits toggle. Therefore, it can be said that it is only not possible to make all digits in binary string equal when the count of both 1 and 0 is odd.

Below is the implementation of the above approach:

## C++

`// C++ program for the` `// above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if` `// Binary string can be` `// made equal` `string canMake(string& s)` `{` ` ` `int` `o = 0, z = 0;` ` ` `// Counting occurence of` ` ` `// zero and one in binary` ` ` `// string` ` ` `for` `(` `int` `i = 0; i < s.size(); i++) {` ` ` `if` `(s[i] - ` `'0'` `== 1)` ` ` `o++;` ` ` `else` ` ` `z++;` ` ` `}` ` ` `// From above observation` ` ` `if` `(o % 2 == 1 && z % 2 == 1)` ` ` `return` `"NO"` `;` ` ` `else` ` ` `return` `"YES"` `;` `}` `// Driver code` `int` `main()` `{` ` ` `string s = ` `"01011"` `;` ` ` `cout << canMake(s) << ` `'\n'` `;` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `class` `GFG` `{` ` ` ` ` `// Function to check if` ` ` `// Binary string can be` ` ` `// made equal` ` ` `static` `String canMake(String s)` ` ` `{` ` ` ` ` `int` `o = ` `0` `, z = ` `0` `;` ` ` ` ` `// Counting occurence of` ` ` `// zero and one in binary` ` ` `// string` ` ` `for` `(` `int` `i = ` `0` `; i < s.length(); i++)` ` ` `{` ` ` `if` `(s.charAt(i) - ` `'0'` `== ` `1` `)` ` ` `o++;` ` ` `else` ` ` `z++;` ` ` `}` ` ` ` ` `// From above observation` ` ` `if` `(o % ` `2` `== ` `1` `&& z % ` `2` `== ` `1` `)` ` ` `return` `"NO"` `;` ` ` `else` ` ` `return` `"YES"` `;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` ` ` `String s = ` `"01011"` `;` ` ` `System.out.println(canMake(s)) ;` ` ` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Python3

`# Python3 program for the above approach` `# Function to check if` `# Binary string can be` `# made equal` `def` `canMake(s) :` ` ` `o ` `=` `0` `; z ` `=` `0` `;` ` ` `# Counting occurence of` ` ` `# zero and one in binary` ` ` `# string` ` ` `for` `i ` `in` `range` `(` `len` `(s)) :` ` ` `if` `(` `ord` `(s[i]) ` `-` `ord` `(` `'0'` `) ` `=` `=` `1` `) :` ` ` `o ` `+` `=` `1` `;` ` ` `else` `:` ` ` `z ` `+` `=` `1` `;` ` ` `# From above observation` ` ` `if` `(o ` `%` `2` `=` `=` `1` `and` `z ` `%` `2` `=` `=` `1` `) :` ` ` `return` `"NO"` `;` ` ` `else` `:` ` ` `return` `"YES"` `;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `s ` `=` `"01011"` `;` ` ` `print` `(canMake(s));` `# This code is contributed by AnkitRai01` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Function to check if` ` ` `// Binary string can be` ` ` `// made equal` ` ` `static` `string` `canMake(` `string` `s)` ` ` `{` ` ` ` ` `int` `o = 0, z = 0;` ` ` ` ` `// Counting occurence of` ` ` `// zero and one in binary` ` ` `// string` ` ` `for` `(` `int` `i = 0; i < s.Length; i++)` ` ` `{` ` ` `if` `(s[i] - ` `'0'` `== 1)` ` ` `o++;` ` ` `else` ` ` `z++;` ` ` `}` ` ` ` ` `// From above observation` ` ` `if` `(o % 2 == 1 && z % 2 == 1)` ` ` `return` `"NO"` `;` ` ` `else` ` ` `return` `"YES"` `;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `string` `s = ` `"01011"` `;` ` ` `Console.WriteLine(canMake(s)) ;` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// javascript program for the above approach` `// Function to check if` `// Binary string can be` `// made equal` `function` `canMake(s)` `{` ` ` `var` `o = 0, z = 0;` ` ` `// Counting occurence of` ` ` `// zero and one in binary` ` ` `// string` ` ` `for` `(i = 0; i < s.length; i++)` ` ` `{` ` ` `if` `(s.charAt(i).charCodeAt(0) - ` `'0'` `.charCodeAt(0) == 1)` ` ` `o++;` ` ` `else` ` ` `z++;` ` ` `}` ` ` `// From above observation` ` ` `if` `(o % 2 == 1 && z % 2 == 1)` ` ` `return` `"NO"` `;` ` ` `else` ` ` `return` `"YES"` `;` `}` `// Driver code` `var` `s = ` `"01011"` `;` `document.write(canMake(s)) ;` `// This code is contributed by Rajput-Ji` `</script>` |

**Output:**

YES

**Time Complexity:** O(n), where n is the length of the given Binary number

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