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Number of Steps to Reduce a Number to Zero - Bit Manipulation - Easy - LeetCode

Number of Steps to Reduce a Number to Zero - Bit Manipulation - Easy - LeetCode

Given a non-negative integer num, return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

 

Example 1:

Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7. 
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3. 
Step 4) 3 is odd; subtract 1 and obtain 2. 
Step 5) 2 is even; divide by 2 and obtain 1. 
Step 6) 1 is odd; subtract 1 and obtain 0.
Example 2:

Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4. 
Step 2) 4 is even; divide by 2 and obtain 2. 
Step 3) 2 is even; divide by 2 and obtain 1. 
Step 4) 1 is odd; subtract 1 and obtain 0.
Example 3:

Input: num = 123
Output: 12
 

Constraints:

0 <= num <= 10^6

public class Solution {
    public int NumberOfSteps (int num) {
        if(num==0)
        {
            return 0;
        }
        
        int count=0;
        bool isLastOne = false;
        for(int i=0;i<32 && num !=0;i++){
            if((num&1)==0){
                //divide by zero
                count+=1;
            }
            else{
                //substract one and divide by zero
                count+=2;
            }
            isLastOne = (num&1)==1;
            num=num>>1;
        }
        
        //subtracting 1 in both case of MSB as 1 and 0
        count-=1;
        return count;
    }
}

Time Complexity: O(logn) which corresponds to the number of binary bits.

Space Complexity: O(1)