LeetCode #496 — EASY

Next Greater Element I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

Constraints:

1 <= nums1.length <= nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 10⁴
All integers in nums1 and nums2 are unique.
All the integers of nums1 also appear in nums2.

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Step 01

Brute Force Baseline

Problem summary: The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2. For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1. Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

Baseline thinking

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

Pattern signal: Array · Hash Map · Stack

Example 1

[4,1,2]
[1,3,4,2]

Example 2

[2,4]
[1,2,3,4]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #496: Next Greater Element I
class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Deque<Integer> stk = new ArrayDeque<>();
        int m = nums1.length, n = nums2.length;
        Map<Integer, Integer> d = new HashMap(n);
        for (int i = n - 1; i >= 0; --i) {
            int x = nums2[i];
            while (!stk.isEmpty() && stk.peek() < x) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                d.put(x, stk.peek());
            }
            stk.push(x);
        }
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = d.getOrDefault(nums1[i], -1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #496: Next Greater Element I
func nextGreaterElement(nums1 []int, nums2 []int) (ans []int) {
	stk := []int{}
	d := map[int]int{}
	for i := len(nums2) - 1; i >= 0; i-- {
		x := nums2[i]
		for len(stk) > 0 && stk[len(stk)-1] < x {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			d[x] = stk[len(stk)-1]
		}
		stk = append(stk, x)
	}
	for _, x := range nums1 {
		if v, ok := d[x]; ok {
			ans = append(ans, v)
		} else {
			ans = append(ans, -1)
		}
	}
	return
}

# Accepted solution for LeetCode #496: Next Greater Element I
class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        stk = []
        d = {}
        for x in nums2[::-1]:
            while stk and stk[-1] < x:
                stk.pop()
            if stk:
                d[x] = stk[-1]
            stk.append(x)
        return [d.get(x, -1) for x in nums1]

// Accepted solution for LeetCode #496: Next Greater Element I
use std::collections::HashMap;

impl Solution {
    pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let mut stk = Vec::new();
        let mut d = HashMap::new();
        for &x in nums2.iter().rev() {
            while let Some(&top) = stk.last() {
                if top <= x {
                    stk.pop();
                } else {
                    break;
                }
            }
            if let Some(&top) = stk.last() {
                d.insert(x, top);
            }
            stk.push(x);
        }

        nums1
            .into_iter()
            .map(|x| *d.get(&x).unwrap_or(&-1))
            .collect()
    }
}

// Accepted solution for LeetCode #496: Next Greater Element I
function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
    const stk: number[] = [];
    const d: Record<number, number> = {};
    for (const x of nums2.reverse()) {
        while (stk.length && stk.at(-1)! < x) {
            stk.pop();
        }
        d[x] = stk.length ? stk.at(-1)! : -1;
        stk.push(x);
    }
    return nums1.map(x => d[x]);
}

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(m + n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.