Minimum Operations to Make the Integer Zero

Minimum Operations to Make the Integer Zero - Problem

You're tasked with a fascinating mathematical puzzle involving binary manipulation and strategic thinking!

Given two integers num1 and num2, your goal is to reduce num1 to exactly 0 using a specific operation.

The Operation: In each step, you can choose any integer i in the range [0, 60] and subtract 2ⁱ + num2 from num1.

For example, if you choose i = 3, you subtract 2³ + num2 = 8 + num2 from num1.

Your mission: Find the minimum number of operations needed to make num1 equal to 0. If it's impossible, return -1.

This problem combines bit manipulation with mathematical reasoning - you'll need to think about powers of 2 and how they can be strategically combined!

Input & Output

example_1.py — Basic Case

$ Input: num1 = 11, num2 = 1

› Output: 3

💡 Note: We need 3 operations: 11 - (2³ + 1) = 2, then 2 - (2¹ + 1) = -1, then -1 - (2⁰ + 1) = -3. Wait, that's wrong. Let me recalculate: We need target = 11 - 3×1 = 8. Since 8 = 2³, we need exactly 1 power of 2, but we need 3 powers total. We can split: 8 = 4 + 2 + 2 = 2² + 2¹ + 2¹. Actually 8 = 2³ can become 2² + 2² = 4 + 4, then 2² + 2¹ + 2¹ = 4 + 2 + 2, then 2¹ + 2¹ + 2¹ + 2¹ = 2 + 2 + 2 + 2. So 8 can be expressed as sum of 3 powers: 2² + 2¹ + 2¹.

example_2.py — Simple Case

$ Input: num1 = 5, num2 = 7

› Output: -1

💡 Note: Let's check: for k=1, target = 5-7 = -2 (invalid). For k=2, target = 5-14 = -9 (invalid). All targets are negative, so impossible.

example_3.py — Edge Case

$ Input: num1 = 1000000000, num2 = 0

› Output: 30

💡 Note: With num2=0, we need sum of k powers of 2 = 1000000000. Since 1000000000 in binary has exactly 13 bits set, we need at least 13 operations. But we can use more by splitting powers. The answer depends on finding the minimum k where popcount(1000000000) ≤ k ≤ 1000000000.

Visualization

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Understanding the Visualization

Setup

We have num1 tokens to spend, each operation costs 2^i + num2

Analysis

After k operations: total spent = k×num2 + sum of k powers of 2

Target

We need: sum of k powers of 2 = num1 - k×num2 = target

Validation

Check if target can be expressed as exactly k powers of 2

Key Takeaway

🎯 Key Insight: The problem reduces to checking if a number can be expressed as a sum of exactly k distinct powers of 2, which is determined by its binary representation and the constraint that we can "split" larger powers into smaller ones.

Time & Space Complexity

Time Complexity

⏱️

O(60 * 60)

We try up to 60 operations, and for each we do constant time checks

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for calculations

✓ Linear Space

Constraints

1 ≤ num1 ≤ 10⁹
-10⁹ ≤ num2 ≤ 10⁹
0 ≤ i ≤ 60 (powers of 2 from 2⁰ to 2⁶⁰)

Asked in

G Google 15 f Meta 8 B ByteDance 6 ⊞ Microsoft 4

The optimal solution uses mathematical analysis combined with bit manipulation. Key insight: after k operations, we need target = num1 - k×num2 to be expressible as a sum of exactly k powers of 2. This is possible when popcount(target) ≤ k ≤ target. The algorithm tries k from 1 to 60 and returns the first valid k.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Enumeration	O(60 * 60)	O(1)	Try all possible combinations of operations up to a reasonable limit
Mathematical Analysis with Binary Search Bounds	O(60)	O(1)	Use mathematical constraints to narrow down the search space efficiently

Brute Force Enumeration — Algorithm Steps

Try k = 1, 2, 3, ... operations up to a reasonable limit
For each k, calculate target = num1 - k * num2
Check if target can be expressed as sum of exactly k distinct powers of 2
Use bit manipulation to verify: target must have exactly k bits set and be >= k
Return the first valid k found, or -1 if none exists

Visualization

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Step-by-Step Walkthrough

Try k=1

Check if num1 - 1*num2 can be expressed as exactly 1 power of 2

Try k=2

Check if num1 - 2*num2 can be expressed as exactly 2 powers of 2

Continue

Keep trying until we find a valid k or exhaust possibilities

Code -

solution.c — C

#include <stdio.h>
#include <stdint.h>

int popcount(long long x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

int makeTheIntegerZero(int num1, int num2) {
    // Try k operations from 1 to 60
    for (int k = 1; k <= 60; k++) {
        // After k operations: num1 - k*num2 should equal sum of k powers of 2
        long long target = (long long)num1 - (long long)k * num2;
        
        // Target must be positive and at least k
        if (target < k) continue;
        
        // Count number of 1-bits in target
        int minPowers = popcount(target);
        
        // Check if we can use exactly k powers of 2
        if (minPowers <= k && k <= target) {
            return k;
        }
    }
    
    return -1;
}

int main() {
    int num1, num2;
    scanf("%d %d", &num1, &num2);
    printf("%d\n", makeTheIntegerZero(num1, num2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(60 * 60)

We try up to 60 operations, and for each we do constant time checks

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for calculations

✓ Linear Space

Constraints

1 ≤ num1 ≤ 10⁹
-10⁹ ≤ num2 ≤ 10⁹
0 ≤ i ≤ 60 (powers of 2 from 2⁰ to 2⁶⁰)

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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