LeetCode #2454 — HARD

Next Greater Element IV

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.

The second greater integer of nums[i] is nums[j] such that:

  • j > i
  • nums[j] > nums[i]
  • There exists exactly one index k such that nums[k] > nums[i] and i < k < j.

If there is no such nums[j], the second greater integer is considered to be -1.

  • For example, in the array [1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3, and that of 3 and 4 is -1.

Return an integer array answer, where answer[i] is the second greater integer of nums[i].

Example 1:

Input: nums = [2,4,0,9,6]
Output: [9,6,6,-1,-1]
Explanation:
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].

Example 2:

Input: nums = [3,3]
Output: [-1,-1]
Explanation:
We return [-1,-1] since neither integer has any integer greater than it.

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
Patterns Used
🔍Modified Binary Search→📚Monotonic Stack→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer. The second greater integer of nums[i] is nums[j] such that: j > i nums[j] > nums[i] There exists exactly one index k such that nums[k] > nums[i] and i < k < j. If there is no such nums[j], the second greater integer is considered to be -1. For example, in the array [1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3, and that of 3 and 4 is -1. Return an integer array answer, where answer[i] is the second greater integer of nums[i].

Baseline thinking

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

Pattern signal: Array · Binary Search · Stack

Example 1

[2,4,0,9,6]

Example 2

[3,3]

Related Problems

  • Next Greater Element I (next-greater-element-i)
  • Replace Elements with Greatest Element on Right Side (replace-elements-with-greatest-element-on-right-side)
  • Apply Operations to Maximize Score (apply-operations-to-maximize-score)
Step 02

Core Insight

What unlocks the optimal approach

  • Move forward in nums and store the value in a non-increasing stack for the first greater value.
  • Move the value in the stack to an ordered data structure for the second greater value.
  • Move value from the ordered data structure for the answer.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #2454: Next Greater Element IV
class Solution {
    public int[] secondGreaterElement(int[] nums) {
        int n = nums.length;
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        int[][] arr = new int[n][0];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {nums[i], i};
        }
        Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        TreeSet<Integer> ts = new TreeSet<>();
        for (int[] pair : arr) {
            int i = pair[1];
            Integer j = ts.higher(i);
            if (j != null && ts.higher(j) != null) {
                ans[i] = nums[ts.higher(j)];
            }
            ts.add(i);
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n)

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.

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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