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:

_Input: __N = 12 _

_Output: __2 _

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: __0 _

Explanation:

_Binary String representing 7 = 111, which is already a palindrome. _

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

**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 first and the second pointer is same or not. If not, increment the number of bits to change.

Below is the implementation of the above approach:

• C++
• Python3

// C++ Program to implement

// the above approach

#include <bits/stdc++.h>

**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;

}

Output:

2

Time Complexity:_ O(log N)_

Auxiliary Space:_ O(1)_

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#bit magic #greedy #mathematical #strings #palindrome