Minimum Bit Flips to Convert Number

#leetcode #bitmaniplulation #dsa

We have to find out how many bits are needed to be flipped in start to make it equal to goal
We know that exor produces 1 for odd number of 1 else 0
This can help us in identifying how many bits are needed to be flipped in start to make it equal to goal

TC: $O(log_{2}X)$ Where X is (start exor end)
SC: O(1)

class Solution {
    public int minBitFlips(int start, int goal) {
        int exorVal  = start ^ goal;

        StringBuilder str = new StringBuilder();
        int count = 0;

//logarithmic time complexity
        while(exorVal!=0){
            if(exorVal%2 ==1) count++;
            exorVal = exorVal/2;
        }
        return count;
    }
}