Coding Interview PatternsNext Greater Element I
EasyStacks
Next Greater Element I
Explanation & Solution
Description
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.
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output:[-1,3,-1]
0
-1
1
3
2
-1
Explanation:
- For 4: no next greater element in nums2 → -1
- For 1: next greater element in nums2 is 3
- For 2: no next greater element in nums2 → -1
Constraints
1 <= nums1.length <= nums2.length <= 10000 <= nums1[i], nums2[i] <= 10^4- All integers in
nums1andnums2are unique - All the integers of
nums1also appear innums2
Approach
Stacks pattern
1. Precompute Next Greater Elements
- Process
nums2with a monotonic decreasing stack - For each element, pop all smaller elements from the stack — the current element is their "next greater"
- Store these mappings in a hash map
2. Build the Stack
- For each element in
nums2: - While the stack is non-empty and the current element is greater than the top:
- Pop the top and record:
map[popped] = current - Push the current element
3. Query for nums1
- For each element in
nums1, look up its next greater element in the map - If not found (element was never popped), return
-1
Key Insight
- The monotonic stack processes each element of
nums2exactly once (push + pop), giving O(n) preprocessing - The hash map provides O(1) lookups for each query in
nums1 - Total time: O(n + m) where n = nums2.length, m = nums1.length
Visualization
Input:
[4, 1, 2]
—
Press ▶ or use ← → to step through
Left (L)Right (R)ConvergedDone
7 steps