# Minimum Count of Bit flips required to make a Binary String Palindromic

• Last Updated : 16 Nov, 2021

Given an integer N, the task is to find the minimum number of bits required to be flipped to convert the binary representation of N into a palindrome.

Examples:

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Input: N = 12
Output:
Explanation:
Binary String representing 12 = “1100”.
To make “1100” a palindrome, convert the string to “0110”.
Therefore, minimum bits required to be flipped is 2.
Input: N = 7
Output:
Explanation:
Binary String representing 7 = 111, which is already a palindrome.

Naive Approach: The simplest way is to check for every possible subset that is a palindrome having the same number of bits.
Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized through these steps:

1. At first, check the length of the binary form of the given number.
2. Take two pointers to one at the L.S.B and another to the M.S.B.
3.  Now keep decrementing the first pointer and incrementing the second pointer.
4. Check whether the bits at both the position of the first and the second pointer are the same or not. If not, increment the number of bits to change.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to calculate the``// length of the binary string``int` `check_length(``int` `n)``{``    ``// Length``    ``int` `ans = 0;``    ``while` `(n) {` `        ``// Right shift of n``        ``n = n >> 1;` `        ``// Increment the length``        ``ans++;``    ``}` `    ``// Return the length``    ``return` `ans;``}` `// Function to check if the bit present``// at i-th position is a set bit or not``int` `check_ith_bit(``int` `n, ``int` `i)``{``    ``// Returns true if the bit is set``    ``return` `(n & (1 << (i - 1)))``               ``? ``true``               ``: ``false``;``}` `// Function to count the minimum``// number of bit flips required``int` `no_of_flips(``int` `n)``{``    ``// Length of the binary form``    ``int` `len = check_length(n);` `    ``// Number of flips``    ``int` `ans = 0;` `    ``// Pointer to the LSB``    ``int` `right = 1;` `    ``// Pointer to the MSB``    ``int` `left = len;` `    ``while` `(right < left) {` `        ``// Check if the bits are equal``        ``if` `(check_ith_bit(n, right)``            ``!= check_ith_bit(n, left))``            ``ans++;` `        ``// Decrementing the``        ``// left pointer``        ``left--;` `        ``// Incrementing the``        ``// right pointer``        ``right++;``    ``}` `    ``// Returns the number of``    ``// bits to flip.``    ``return` `ans;``}` `// Driver Code``int` `main()``{` `    ``int` `n = 12;` `    ``cout << no_of_flips(n);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``class` `GFG{` `// Function to calculate the``// length of the binary string``static` `int` `check_length(``int` `n)``{``    ` `    ``// Length``    ``int` `ans = ``0``;``    ` `    ``while` `(n != ``0``)``    ``{` `        ``// Right shift of n``        ``n = n >> ``1``;` `        ``// Increment the length``        ``ans++;``    ``}` `    ``// Return the length``    ``return` `ans;``}` `// Function to check if the bit present``// at i-th position is a set bit or not``static` `boolean` `check_ith_bit(``int` `n, ``int` `i)``{``    ` `    ``// Returns true if the bit is set``    ``return` `(n & (``1``<< (i - ``1``))) != ``0` `? ``true` `: ``false``;``}` `// Function to count the minimum``// number of bit flips required``static` `int` `no_of_flips(``int` `n)``{``    ` `    ``// Length of the binary form``    ``int` `len = check_length(n);` `    ``// Number of flips``    ``int` `ans = ``0``;` `    ``// Pointer to the LSB``    ``int` `right = ``1``;` `    ``// Pointer to the MSB``    ``int` `left = len;` `    ``while` `(right < left)``    ``{``        ` `        ``// Check if the bits are equal``        ``if` `(check_ith_bit(n, right) !=``            ``check_ith_bit(n, left))``            ``ans++;` `        ``// Decrementing the``        ``// left pointer``        ``left--;` `        ``// Incrementing the``        ``// right pointer``        ``right++;``    ``}` `    ``// Returns the number of``    ``// bits to flip.``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``12``;` `    ``System.out.println(no_of_flips(n));``}``}` `// This code is contributed by rutvik_56`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to calculate the``# length of the binary string``def` `check_length(n):` `    ``# Length``    ``ans ``=` `0``    ` `    ``while` `(n):` `        ``# Right shift of n``        ``n ``=` `n >> ``1` `        ``# Increment the length``        ``ans ``+``=` `1` `    ``# Return the length``    ``return` `ans` `# Function to check if the bit present``# at i-th position is a set bit or not``def` `check_ith_bit(n, i):` `    ``# Returns true if the bit is set``    ``if` `(n & (``1` `<< (i ``-` `1``))):``        ``return` `True``    ``else``:``        ``return` `False` `# Function to count the minimum``# number of bit flips required``def` `no_of_flips(n):` `    ``# Length of the binary form``    ``ln ``=` `check_length(n)` `    ``# Number of flips``    ``ans ``=` `0` `    ``# Pointer to the LSB``    ``right ``=` `1` `    ``# Pointer to the MSB``    ``left ``=` `ln` `    ``while` `(right < left):` `        ``# Check if the bits are equal``        ``if` `(check_ith_bit(n, right) !``=``            ``check_ith_bit(n, left)):``            ``ans ``+``=` `1` `        ``# Decrementing the``        ``# left pointer``        ``left ``-``=` `1` `        ``# Incrementing the``        ``# right pointer``        ``right ``+``=` `1` `    ``# Returns the number of``    ``# bits to flip.``    ``return` `ans` `# Driver Code``n ``=` `12` `print``(no_of_flips(n))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `// Function to calculate the``// length of the binary string``static` `int` `check_length(``int` `n)``{``    ` `    ``// Length``    ``int` `ans = 0;``    ` `    ``while` `(n != 0)``    ``{` `        ``// Right shift of n``        ``n = n >> 1;` `        ``// Increment the length``        ``ans++;``    ``}` `    ``// Return the length``    ``return` `ans;``}` `// Function to check if the bit present``// at i-th position is a set bit or not``static` `bool` `check_ith_bit(``int` `n, ``int` `i)``{``    ` `    ``// Returns true if the bit is set``    ``return` `(n & (1 << (i - 1))) != 0 ?``                                ``true` `: ``false``;``}` `// Function to count the minimum``// number of bit flips required``static` `int` `no_of_flips(``int` `n)``{``    ` `    ``// Length of the binary form``    ``int` `len = check_length(n);` `    ``// Number of flips``    ``int` `ans = 0;` `    ``// Pointer to the LSB``    ``int` `right = 1;` `    ``// Pointer to the MSB``    ``int` `left = len;` `    ``while` `(right < left)``    ``{``        ` `        ``// Check if the bits are equal``        ``if` `(check_ith_bit(n, right) !=``            ``check_ith_bit(n, left))``            ``ans++;` `        ``// Decrementing the``        ``// left pointer``        ``left--;` `        ``// Incrementing the``        ``// right pointer``        ``right++;``    ``}` `    ``// Returns the number of``    ``// bits to flip.``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 12;` `    ``Console.WriteLine(no_of_flips(n));``}``}` `// This code is contributed by sapnasingh4991`

## Javascript

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Output:
`2`

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

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