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.

Thinking Process

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.

  • Stack matches nested or LIFO structure (parentheses, monotonic scans).
  • Push on open / larger; pop when the current element resolves pending work.
  • Monotonic stack finds next greater/smaller in O(n).
Stack top push / pop LIFO — monotonic stack scans array

Common Approaches

Typical techniques for this pattern:

Approach Time Space Notes
Monotonic stack (this problem) O(n) O(n) Next greater/smaller element
Parentheses matching O(n) O(n) Push open, pop on close
Expression evaluation O(n) O(n) Operand + operator stacks
Stack simulation O(n) O(n) Process in LIFO order

Examples

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] <= 10^4
  • All integers in nums1 and nums2 are unique.
  • All the integers of nums1 also appear in nums2.

Algorithm Breakdown

Key Insight: Monotonic Stack

The stack maintains elements in decreasing order (from bottom to top):

  • When we see a new element, pop all smaller or equal elements
  • The top of stack is the next greater element
  • Push current element to maintain monotonic property

Right-to-Left Traversal

Processing from right to left ensures:

  • We’ve already processed elements to the right
  • Stack contains potential next greater elements
  • Each element’s next greater is already in stack

Why This Works

  1. Stack Property: Stack stores elements in decreasing order
  2. Popping Logic: Elements ≤ current can’t be next greater for anything left
  3. Top Element: Top of stack is the first greater element to the right
  4. Hash Map: Stores results for O(1) lookup later

Monotonic Stack Template

Here’s the general template for next greater element problems:

vector<int> nextGreaterElement(vector<int>& nums) {
    int n = nums.size();
    vector<int> result(n, -1);
    stack<int> stk;
    
    // Traverse from right to left
    for(int i = n - 1; i >= 0; i--) {
        // Pop elements that can't be next greater
        while(!stk.empty() && nums[i] >= stk.top()) {
            stk.pop();
        }
        
        // Top of stack is next greater element
        if(!stk.empty()) {
            result[i] = stk.top();
        }
        
        // Push current element
        stk.push(nums[i]);
    }
    
    return result;
}

Key Template Components:

  1. Right-to-Left Traversal: Process array backwards
  2. Monotonic Stack: Maintain decreasing order
  3. Pop Condition: Remove elements ≤ current
  4. Result Storage: Store next greater or -1

Complexity

Time Complexity: O(n + m)

  • Process nums2: O(n) where n = nums2.length
  • Query nums1: O(m) where m = nums1.length
  • Total: O(n + m)

Space Complexity: O(n)

  • Hash map: O(n) - stores next greater for each element in nums2
  • Stack: O(n) - worst case all elements in stack
  • Total: O(n)

Key Points

  1. Monotonic Stack: Maintains elements in decreasing order
  2. Right-to-Left: Process from end to find next greater efficiently
  3. Hash Map: O(1) lookup for results
  4. Single Pass: Process nums2 once, query nums1 once
  5. Efficient: O(n + m) time complexity

Detailed Example Walkthrough

Example: nums1 = [2,4], nums2 = [1,2,3,4]

Step 1: Process nums2 from right to left

i=3: num = 4
  Stack: []
  Pop: nothing
  hashmap[4] = -1
  Stack: [4]

i=2: num = 3
  Stack: [4]
  Pop: 4 <= 3? No → keep 4
  hashmap[3] = 4
  Stack: [3, 4]

i=1: num = 2
  Stack: [3, 4]
  Pop: 3 <= 2? No → keep 3
  hashmap[2] = 3
  Stack: [2, 3, 4]

i=0: num = 1
  Stack: [2, 3, 4]
  Pop: 2 <= 1? No → keep 2
  hashmap[1] = 2
  Stack: [1, 2, 3, 4]

hashmap = {4: -1, 3: 4, 2: 3, 1: 2}

Step 2: Query nums1
nums1[0] = 2 → hashmap[2] = 3
nums1[1] = 4 → hashmap[4] = -1

Result: [3, -1]

Edge Cases

  1. No greater element: Element is maximum in nums2 → return -1
  2. Single element: nums2 has one element → return -1
  3. All decreasing: nums2 in decreasing order → all return -1
  4. All increasing: nums2 in increasing order → each has next greater
  5. Duplicate handling: Problem states all integers are unique

Common Mistakes

  • Skipping edge cases (empty input, single element, boundaries).
  • Off-by-one errors in loops and index ranges.
  • Forgetting to handle the case when no valid answer exists.

Tags

Array, Stack, Monotonic Stack, Hash Table, Easy

Key Takeaways

  • Stack matches nested or LIFO structure (parentheses, monotonic scans).
  • Push on open / larger; pop when the current element resolves pending work.
  • Monotonic stack finds next greater/smaller in O(n).

References

Template Reference