LeetCode #1860 — MEDIUM

Incremental Memory Leak

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

The Problem

Problem Statement

You are given two integers memory1 and memory2 representing the available memory in bits on two memory sticks. There is currently a faulty program running that consumes an increasing amount of memory every second.

At the i^th second (starting from 1), i bits of memory are allocated to the stick with more available memory (or from the first memory stick if both have the same available memory). If neither stick has at least i bits of available memory, the program crashes.

Return an array containing [crashTime, memory1_crash, memory2_crash], where crashTime is the time (in seconds) when the program crashed and memory1_crash and memory2_crash are the available bits of memory in the first and second sticks respectively.

Example 1:

Input: memory1 = 2, memory2 = 2
Output: [3,1,0]
Explanation: The memory is allocated as follows:
- At the 1^st second, 1 bit of memory is allocated to stick 1. The first stick now has 1 bit of available memory.
- At the 2^nd second, 2 bits of memory are allocated to stick 2. The second stick now has 0 bits of available memory.
- At the 3^rd second, the program crashes. The sticks have 1 and 0 bits available respectively.

Example 2:

Input: memory1 = 8, memory2 = 11
Output: [6,0,4]
Explanation: The memory is allocated as follows:
- At the 1^st second, 1 bit of memory is allocated to stick 2. The second stick now has 10 bit of available memory.
- At the 2^nd second, 2 bits of memory are allocated to stick 2. The second stick now has 8 bits of available memory.
- At the 3^rd second, 3 bits of memory are allocated to stick 1. The first stick now has 5 bits of available memory.
- At the 4^th second, 4 bits of memory are allocated to stick 2. The second stick now has 4 bits of available memory.
- At the 5^th second, 5 bits of memory are allocated to stick 1. The first stick now has 0 bits of available memory.
- At the 6^th second, the program crashes. The sticks have 0 and 4 bits available respectively.

Constraints:

0 <= memory1, memory2 <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: You are given two integers memory1 and memory2 representing the available memory in bits on two memory sticks. There is currently a faulty program running that consumes an increasing amount of memory every second. At the ith second (starting from 1), i bits of memory are allocated to the stick with more available memory (or from the first memory stick if both have the same available memory). If neither stick has at least i bits of available memory, the program crashes. Return an array containing [crashTime, memory1crash, memory2crash], where crashTime is the time (in seconds) when the program crashed and memory1crash and memory2crash are the available bits of memory in the first and second sticks respectively.

Baseline thinking

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

Pattern signal: Math

Example 1

2
2

Example 2

8
11

Step 02

Core Insight

What unlocks the optimal approach

What is the upper bound for the number of seconds?
Simulate the process of allocating memory.

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 #1860: Incremental Memory Leak
class Solution {
    public int[] memLeak(int memory1, int memory2) {
        int i = 1;
        for (; i <= Math.max(memory1, memory2); ++i) {
            if (memory1 >= memory2) {
                memory1 -= i;
            } else {
                memory2 -= i;
            }
        }
        return new int[] {i, memory1, memory2};
    }
}

// Accepted solution for LeetCode #1860: Incremental Memory Leak
func memLeak(memory1 int, memory2 int) []int {
	i := 1
	for ; i <= memory1 || i <= memory2; i++ {
		if memory1 >= memory2 {
			memory1 -= i
		} else {
			memory2 -= i
		}
	}
	return []int{i, memory1, memory2}
}

# Accepted solution for LeetCode #1860: Incremental Memory Leak
class Solution:
    def memLeak(self, memory1: int, memory2: int) -> List[int]:
        i = 1
        while i <= max(memory1, memory2):
            if memory1 >= memory2:
                memory1 -= i
            else:
                memory2 -= i
            i += 1
        return [i, memory1, memory2]

// Accepted solution for LeetCode #1860: Incremental Memory Leak
/**
 * [1860] Incremental Memory Leak
 *
 * You are given two integers memory1 and memory2 representing the available memory in bits on two memory sticks. There is currently a faulty program running that consumes an increasing amount of memory every second.
 * At the i^th second (starting from 1), i bits of memory are allocated to the stick with more available memory (or from the first memory stick if both have the same available memory). If neither stick has at least i bits of available memory, the program crashes.
 * Return an array containing [crashTime, memory1crash, memory2crash], where crashTime is the time (in seconds) when the program crashed and memory1crash and memory2crash are the available bits of memory in the first and second sticks respectively.
 *  
 * Example 1:
 *
 * Input: memory1 = 2, memory2 = 2
 * Output: [3,1,0]
 * Explanation: The memory is allocated as follows:
 * - At the 1^st second, 1 bit of memory is allocated to stick 1. The first stick now has 1 bit of available memory.
 * - At the 2^nd second, 2 bits of memory are allocated to stick 2. The second stick now has 0 bits of available memory.
 * - At the 3^rd second, the program crashes. The sticks have 1 and 0 bits available respectively.
 *
 * Example 2:
 *
 * Input: memory1 = 8, memory2 = 11
 * Output: [6,0,4]
 * Explanation: The memory is allocated as follows:
 * - At the 1^st second, 1 bit of memory is allocated to stick 2. The second stick now has 10 bit of available memory.
 * - At the 2^nd second, 2 bits of memory are allocated to stick 2. The second stick now has 8 bits of available memory.
 * - At the 3^rd second, 3 bits of memory are allocated to stick 1. The first stick now has 5 bits of available memory.
 * - At the 4^th second, 4 bits of memory are allocated to stick 2. The second stick now has 4 bits of available memory.
 * - At the 5^th second, 5 bits of memory are allocated to stick 1. The first stick now has 0 bits of available memory.
 * - At the 6^th second, the program crashes. The sticks have 0 and 4 bits available respectively.
 *
 *  
 * Constraints:
 *
 * 	0 <= memory1, memory2 <= 2^31 - 1
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/incremental-memory-leak/
// discuss: https://leetcode.com/problems/incremental-memory-leak/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn mem_leak(memory1: i32, memory2: i32) -> Vec<i32> {
        let mut memory1 = memory1;
        let mut memory2 = memory2;
        let mut i = 0;

        loop {
            i += 1;

            if memory1 >= memory2 {
                if memory1 < i {
                    break;
                }
                memory1 -= i;
            } else {
                if memory2 < i {
                    break;
                }
                memory2 -= i;
            }
        }

        vec![i, memory1, memory2]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1860_example_1() {
        let memory1 = 2;
        let memory2 = 2;

        let result = vec![3, 1, 0];

        assert_eq!(Solution::mem_leak(memory1, memory2), result);
    }

    #[test]
    fn test_1860_example_2() {
        let memory1 = 8;
        let memory2 = 11;

        let result = vec![6, 0, 4];

        assert_eq!(Solution::mem_leak(memory1, memory2), result);
    }
}

// Accepted solution for LeetCode #1860: Incremental Memory Leak
function memLeak(memory1: number, memory2: number): number[] {
    let i = 1;
    for (; i <= Math.max(memory1, memory2); ++i) {
        if (memory1 >= memory2) {
            memory1 -= i;
        } else {
            memory2 -= i;
        }
    }
    return [i, memory1, memory2];
}

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 n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.