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Difficulty: 🟡 Medium | Topics: Binary Search · Math · Greedy
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] × 1 + workerTimes[i] × 2 + ... + workerTimes[i] × x
= workerTimes[i] × (x × (x + 1) / 2) seconds
🎯 Return the minimum number of seconds required for all workers to reduce the height to 0.
Input : mountainHeight = 4, workerTimes = [2, 1, 1]
Output: 3
| Worker | Height Reduced | Time Taken |
|---|---|---|
| 0 | 1 | 2s |
| 1 | 2 | 1 + 2 = 3s |
| 2 | 1 | 1s |
✅ max(2, 3, 1) = 3 seconds
Input : mountainHeight = 10, workerTimes = [3, 2, 2, 4]
Output: 12
| Worker | Height Reduced | Time Taken |
|---|---|---|
| 0 | 2 | 3 + 6 = 9s |
| 1 | 3 | 2 + 4 + 6 = 12s |
| 2 | 3 | 2 + 4 + 6 = 12s |
| 3 | 2 | 4 + 8 = 12s |
✅ max(9, 12, 12, 12) = 12 seconds
Input : mountainHeight = 5, workerTimes = [1]
Output: 15
| Worker | Height Reduced | Time Taken |
|---|---|---|
| 0 | 5 | 1+2+3+4+5 = 15s |
✅ 15 seconds
The cost for a worker with time w to reduce height by x:
cost(w, x) = w × x × (x + 1) / 2
Given a time budget T, max height a worker can reduce:
solve: w × x × (x + 1) / 2 ≤ T
→ x × (x + 1) ≤ 2T / w
→ x ≈ floor( (-1 + sqrt(1 + 8T/w)) / 2 )
1. Binary search on time T in range [0, max_cost]
2. For each T, greedily compute max height each worker can reduce
3. Check if total reductions ≥ mountainHeight
4. Minimize T
class Solution {
public long minNumberOfSeconds(int mountainHeight, int[] workerTimes) {
// ✅ Compute tight upper bound: worst worker does all the work
long maxTime = Long.MIN_VALUE;
for (int w : workerTimes) {
long h = mountainHeight;
long cost = (long) w * h * (h + 1) / 2;
maxTime = Math.max(maxTime, cost);
}
long lo = 0, hi = maxTime;
while (lo < hi) {
long mid = lo + (hi - lo) / 2;
if (canFinish(mid, mountainHeight, workerTimes))
hi = mid;
else
lo = mid + 1;
}
return lo;
}
private boolean canFinish(long time, int height, int[] workerTimes) {
long total = 0;
for (int w : workerTimes) {
long x = (long)((-1 + Math.sqrt(1 + 8.0 * time / w)) / 2);
// ✅ Correct floating point rounding errors
while ((long) w * (x + 1) * (x + 2) / 2 <= time) x++;
while (x > 0 && (long) w * x * (x + 1) / 2 > time) x--;
total += x;
if (total >= height) return true;
}
return total >= height;
}
}
| Complexity | |
|---|---|
| Time | O(n × log(w × H²)) |
| Space | O(1) |
Where maxTime ≈ w × H² ≈ 10^16 → log factor ≈ 53 iterations.
n = number of workersw = max value in workerTimes (up to 10⁶)H² = shorthand for H × (H+1) / 2 where H = mountainHeight (up to 10⁵) — search range scales quadratically with heightLoading component...
Follow every state change, comparison, and transformation as the execution unfolds in real time, so you understand not just the result, but the journey.
Follow every state change, comparison, and transformation as the execution unfolds in real time, so you understand not just the result, but the journey.
The algorithm is divided into three logical parts. Carefully rearrange each section in the correct order to form a complete and valid solution.
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