LeetCode #2749 — MEDIUM

Minimum Operations to Make the Integer Zero

Move from brute-force thinking to an efficient approach using bit manipulation strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given two integers num1 and num2.

In one operation, you can choose integer i in the range [0, 60] and subtract 2ⁱ + num2 from num1.

Return the integer denoting the minimum number of operations needed to make num1 equal to 0.

If it is impossible to make num1 equal to 0, return -1.

Example 1:

Input: num1 = 3, num2 = -2
Output: 3
Explanation: We can make 3 equal to 0 with the following operations:
- We choose i = 2 and subtract 2² + (-2) from 3, 3 - (4 + (-2)) = 1.
- We choose i = 2 and subtract 2² + (-2) from 1, 1 - (4 + (-2)) = -1.
- We choose i = 0 and subtract 2⁰ + (-2) from -1, (-1) - (1 + (-2)) = 0.
It can be proven, that 3 is the minimum number of operations that we need to perform.

Example 2:

Input: num1 = 5, num2 = 7
Output: -1
Explanation: It can be proven, that it is impossible to make 5 equal to 0 with the given operation.

Constraints:

1 <= num1 <= 10⁹
-10⁹ <= num2 <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given two integers num1 and num2. In one operation, you can choose integer i in the range [0, 60] and subtract 2i + num2 from num1. Return the integer denoting the minimum number of operations needed to make num1 equal to 0. If it is impossible to make num1 equal to 0, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Bit Manipulation

Example 1

3
-2

Example 2

5
7

Core Insight

What unlocks the optimal approach

If we want to make integer n equal to 0 by only subtracting powers of 2 from n, in how many operations can we achieve it?
We need at least - the number of bits in the binary representation of n, and at most - n.
Notice that, if it is possible to make num1 equal to 0, then we need at most 60 operations.
Iterate on the number of operations.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2749: Minimum Operations to Make the Integer Zero
class Solution {
    public int makeTheIntegerZero(int num1, int num2) {
        for (long k = 1;; ++k) {
            long x = num1 - k * num2;
            if (x < 0) {
                break;
            }
            if (Long.bitCount(x) <= k && k <= x) {
                return (int) k;
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #2749: Minimum Operations to Make the Integer Zero
func makeTheIntegerZero(num1 int, num2 int) int {
	for k := 1; ; k++ {
		x := num1 - k*num2
		if x < 0 {
			break
		}
		if bits.OnesCount(uint(x)) <= k && k <= x {
			return k
		}
	}
	return -1
}

# Accepted solution for LeetCode #2749: Minimum Operations to Make the Integer Zero
class Solution:
    def makeTheIntegerZero(self, num1: int, num2: int) -> int:
        for k in count(1):
            x = num1 - k * num2
            if x < 0:
                break
            if x.bit_count() <= k <= x:
                return k
        return -1

// Accepted solution for LeetCode #2749: Minimum Operations to Make the Integer Zero
impl Solution {
    pub fn make_the_integer_zero(num1: i32, num2: i32) -> i32 {
        let num1 = num1 as i64;
        let num2 = num2 as i64;
        for k in 1.. {
            let x = num1 - k * num2;
            if x < 0 {
                break;
            }
            if (x.count_ones() as i64) <= k && k <= x {
                return k as i32;
            }
        }
        -1
    }
}

// Accepted solution for LeetCode #2749: Minimum Operations to Make the Integer Zero
function makeTheIntegerZero(num1: number, num2: number): number {
    for (let k = 1; ; ++k) {
        let x = num1 - k * num2;
        if (x < 0) {
            break;
        }
        if (x.toString(2).replace(/0/g, '').length <= k && k <= x) {
            return k;
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log x)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.