LeetCode - next-greater-element-i challenge solution

Sat Jul 11 2020

This question was quite tricky, I would not take credit on the cool solution as I solved it only with the brute force way. But after reading the stack solution, I went ahead and implemented it myself.

The exercise here is that you get two arrays A & B. For each item in A, you need to find it on B and then find the closest bigger element on its right. If you could not find a bigger element on the right the solution for this element will be -1.

In the end you need to return a new array of the found bigger elements and -1's. Example:

A = [1, 2, 3]
B = [4, 1, 0, 2, 3, -1]

Lets start finding the elements:

  • For A[0], 1 we find it on B[1]. The first bigger elemnt on its right is B[3], 2.
  • For A[1], 2 we find it on B[3]. The first bigger elemnt on its right is B[4], 3.
  • For A[2], 3 we find it on B[4]. It has no bigger elements on its right, so -1.
  • The solution: [2, 3, -1]

Now I'll share the two solutions.

Brute Force

  • iterate on array A
  • for each element a
    • find its index on array B
    • from that index and further you iterate on array B items
    • for each item on B, b
      • check if b > a
      • if so save b for our final array
      • if we finished our iterations and found nothing, let's save -1 for our final solution
  • finish our iterations, the result is our saved array with the found items.

As you can see in this solution on the worst case scenarion we will be O(n^2)

The code:

def nextGreaterElement(nums1: List[int], nums2: List[int]) -> List[int]:
    nextGreater = []
    idx = 0
    while idx < len(nums1):
        num1 = nums1[idx]
        found = False
        secIdx = nums2.index(num1) + 1
        while secIdx < len(nums2):
            num2 = nums2[secIdx]
            secIdx += 1
            if num2 > num1:
                found = True
        idx += 1
        if not found:
    return nextGreater


  • create a stack s
  • create mappings dict m
  • for b in array B
    • if s is empty, insert into it and continue the loop
    • if head of s is bigger than b
      • push b to s and continue the loop
    • start popping items from s, as long as they are smaller than b
      • for each popped item p
      • set its mapping on m
      • m[p] = b
    • as we finished with the popped items, we insert b into s
  • now we just need to go over the items left in s
    • as for each one of them we could not find a bigger element on its right
    • so we set its mapping to -1

The code:

class Solution:
    def nextGreaterElement(nums1: List[int], nums2: List[int]) -> List[int]:
        mappings = {}
        stack = deque()

        for num in nums2:
            if not stack:
            if stack[-1] > num:
            while stack and stack[-1] < num:
                item = stack.pop()
                mappings[item] = num

        while stack:
            item = stack.pop()
            mappings[item] = -1

        return list(map(lambda x: mappings[x], nums1))