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:
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;
}
}
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