Stack/Next Greater ElementMedium

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Next Greater Element

Stack·Medium

Given an array of integers nums, return a new array result where result[i] is the first element greater than nums[i] that appears to the right of index i in the original array.

If no such element exists, set result[i] = -1.

The classic approach uses a monotonic decreasing stack of indices: iterate left to right; when the current element is greater than the value at the stack's top index, that index has found its next greater element — pop and record the answer, then push the current index.

Examples

Input: nums = [2,1,2,4,3]

Output: [4,2,4,-1,-1]

For index 0 (val 2): next greater is 4. For index 1 (val 1): next greater is 2. For index 2 (val 2): next greater is 4. For index 3 (val 4): no greater element to the right. For index 4 (val 3): no greater element to the right.

Input: nums = [1,3,2]

Output: [3,-1,-1]

Constraints

1 <= nums.length <= 10^4
-10^9 <= nums[i] <= 10^9

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Tests

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Passed32 ms

First attempt — retry to track metric movement here.

Your approach

Pattern:Stack
Time:O(N)
Space:O(1)

Complete approach — pattern and complexity both named.

Coach

What you got right

You correctly matched the sorted input requirement and implemented a linear-time scan using two pointers.

Where it diverged

No divergence detected. The code correctly aligns with the stated approach.

Next attempt: focus on

Try solving related sliding window problems to build familiarity with two-pointer variants.

Signals

No signals fired — clean run, you stayed in flow.