Number of Steps to Reduce a Number to Zero - Problem

Given an integer num, return the number of steps to reduce it to zero.

In one step, if the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

Input & Output

Example 1 — Basic Case

$ Input: num = 14

› Output: 6

💡 Note: Step-by-step: 14 → 7 → 6 → 3 → 2 → 1 → 0. Total of 6 steps.

Example 2 — Small Number

$ Input: num = 8

› Output: 4

💡 Note: 8 → 4 → 2 → 1 → 0. All even numbers except the last step.

Example 3 — Edge Case

$ Input: num = 1

› Output: 1

💡 Note: Only one step needed: 1 → 0

Constraints

0 ≤ num ≤ 10⁶

Visualization

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Asked in

L LeetCode 15 M Microsoft 8

The key insight is that we can either simulate the process step by step or use bit manipulation to calculate the result directly. The optimal approach uses the formula: count of 1-bits + total bits - 1. Time: O(log n), Space: O(1)

Common Approaches

✓ Bit Manipulation Approach

⏱️ Time: O(log n) Space: O(1)

Each 1-bit requires one subtraction operation plus one division. Each 0-bit only needs division. Total operations = number of 1-bits + total bits - 1.

Simulation Approach

⏱️ Time: O(log n) Space: O(1)

Simulate the process exactly as described: if number is even divide by 2, if odd subtract 1. Count steps until we reach 0.

Bit Manipulation Approach — Algorithm Steps

Step 1: Count the number of 1-bits in binary representation
Step 2: Count total number of bits
Step 3: Return count of 1-bits + total bits - 1

Visualization

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Step-by-Step Walkthrough

Binary Form

Convert number to binary representation

Count Bits

Count 1-bits and total bits

Calculate

Formula: ones + total_bits - 1

Code -

solution.c — C

#include <stdio.h>

int countOnes(int n) {
    int count = 0;
    while (n) {
        count += n & 1;
        n >>= 1;
    }
    return count;
}

int countBits(int n) {
    int count = 0;
    while (n) {
        count++;
        n >>= 1;
    }
    return count;
}

int solution(int num) {
    if (num == 0) return 0;
    
    int ones = countOnes(num);
    int totalBits = countBits(num);
    
    return ones + totalBits - 1;
}

int main() {
    int num;
    scanf("%d", &num);
    int result = solution(num);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Bit counting operations are proportional to number of bits

⚡ Linearithmic

Space Complexity

O(1)

Only using variables for bit counting

✓ Linear Space

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Number of Steps to Reduce a Number to Zero - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Bit Manipulation Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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