LeetCode #3296 — MEDIUM

Minimum Number of Seconds to Make Mountain Height Zero

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer mountainHeight denoting the height of a mountain.

You are also given an integer array workerTimes representing the work time of workers in seconds.

The workers work simultaneously to reduce the height of the mountain. For worker i:

To decrease the mountain's height by x, it takes workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x seconds. For example:
- To reduce the height of the mountain by 1, it takes workerTimes[i] seconds.
- To reduce the height of the mountain by 2, it takes workerTimes[i] + workerTimes[i] * 2 seconds, and so on.

Return an integer representing the minimum number of seconds required for the workers to make the height of the mountain 0.

Example 1:

Input: mountainHeight = 4, workerTimes = [2,1,1]

Output: 3

Explanation:

One way the height of the mountain can be reduced to 0 is:

Worker 0 reduces the height by 1, taking workerTimes[0] = 2 seconds.
Worker 1 reduces the height by 2, taking workerTimes[1] + workerTimes[1] * 2 = 3 seconds.
Worker 2 reduces the height by 1, taking workerTimes[2] = 1 second.

Since they work simultaneously, the minimum time needed is max(2, 3, 1) = 3 seconds.

Example 2:

Input: mountainHeight = 10, workerTimes = [3,2,2,4]

Output: 12

Explanation:

Worker 0 reduces the height by 2, taking workerTimes[0] + workerTimes[0] * 2 = 9 seconds.
Worker 1 reduces the height by 3, taking workerTimes[1] + workerTimes[1] * 2 + workerTimes[1] * 3 = 12 seconds.
Worker 2 reduces the height by 3, taking workerTimes[2] + workerTimes[2] * 2 + workerTimes[2] * 3 = 12 seconds.
Worker 3 reduces the height by 2, taking workerTimes[3] + workerTimes[3] * 2 = 12 seconds.

The number of seconds needed is max(9, 12, 12, 12) = 12 seconds.

Example 3:

Input: mountainHeight = 5, workerTimes = [1]

Output: 15

Explanation:

There is only one worker in this example, so the answer is workerTimes[0] + workerTimes[0] * 2 + workerTimes[0] * 3 + workerTimes[0] * 4 + workerTimes[0] * 5 = 15.

Constraints:

1 <= mountainHeight <= 10⁵
1 <= workerTimes.length <= 10⁴
1 <= workerTimes[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer mountainHeight denoting the height of a mountain. You are also given an integer array workerTimes representing the work time of workers in seconds. The workers work simultaneously to reduce the height of the mountain. For worker i: To decrease the mountain's height by x, it takes workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x seconds. For example: To reduce the height of the mountain by 1, it takes workerTimes[i] seconds. To reduce the height of the mountain by 2, it takes workerTimes[i] + workerTimes[i] * 2 seconds, and so on. Return an integer representing the minimum number of seconds required for the workers to make the height of the mountain 0.

Baseline thinking

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

Pattern signal: Array · Math · Binary Search · Greedy

Example 1

4
[2,1,1]

Example 2

10
[3,2,2,4]

Example 3

5
[1]

Step 02

Core Insight

What unlocks the optimal approach

Can we use binary search to solve this problem?
Do a binary search on the number of seconds to check if it's enough to reduce the mountain height to 0 or less with all workers working simultaneously.

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 #3296: Minimum Number of Seconds to Make Mountain Height Zero
class Solution {
    private int mountainHeight;
    private int[] workerTimes;

    public long minNumberOfSeconds(int mountainHeight, int[] workerTimes) {
        this.mountainHeight = mountainHeight;
        this.workerTimes = workerTimes;
        long l = 1, r = (long) 1e16;
        while (l < r) {
            long mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }

    private boolean check(long t) {
        long h = 0;
        for (int wt : workerTimes) {
            h += (long) (Math.sqrt(t * 2.0 / wt + 0.25) - 0.5);
        }
        return h >= mountainHeight;
    }
}

// Accepted solution for LeetCode #3296: Minimum Number of Seconds to Make Mountain Height Zero
func minNumberOfSeconds(mountainHeight int, workerTimes []int) int64 {
	return int64(sort.Search(1e16, func(t int) bool {
		var h int64
		for _, wt := range workerTimes {
			h += int64(math.Sqrt(float64(t)*2.0/float64(wt)+0.25) - 0.5)
		}
		return h >= int64(mountainHeight)
	}))
}

# Accepted solution for LeetCode #3296: Minimum Number of Seconds to Make Mountain Height Zero
class Solution:
    def minNumberOfSeconds(self, mountainHeight: int, workerTimes: List[int]) -> int:
        def check(t: int) -> bool:
            h = 0
            for wt in workerTimes:
                h += int(sqrt(2 * t / wt + 1 / 4) - 1 / 2)
            return h >= mountainHeight

        return bisect_left(range(10**16), True, key=check)

// Accepted solution for LeetCode #3296: Minimum Number of Seconds to Make Mountain Height Zero
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3296: Minimum Number of Seconds to Make Mountain Height Zero
// class Solution {
//     private int mountainHeight;
//     private int[] workerTimes;
// 
//     public long minNumberOfSeconds(int mountainHeight, int[] workerTimes) {
//         this.mountainHeight = mountainHeight;
//         this.workerTimes = workerTimes;
//         long l = 1, r = (long) 1e16;
//         while (l < r) {
//             long mid = (l + r) >> 1;
//             if (check(mid)) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l;
//     }
// 
//     private boolean check(long t) {
//         long h = 0;
//         for (int wt : workerTimes) {
//             h += (long) (Math.sqrt(t * 2.0 / wt + 0.25) - 0.5);
//         }
//         return h >= mountainHeight;
//     }
// }

// Accepted solution for LeetCode #3296: Minimum Number of Seconds to Make Mountain Height Zero
function minNumberOfSeconds(mountainHeight: number, workerTimes: number[]): number {
    const check = (t: bigint): boolean => {
        let h = BigInt(0);
        for (const wt of workerTimes) {
            h += BigInt(Math.floor(Math.sqrt((Number(t) * 2.0) / wt + 0.25) - 0.5));
        }
        return h >= BigInt(mountainHeight);
    };

    let l = BigInt(1);
    let r = BigInt(1e16);

    while (l < r) {
        const mid = (l + r) >> BigInt(1);
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1n;
        }
    }

    return Number(l);
}

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

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.