Leetcode 0496. Next Greater Element I

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`.

Programming Skills I Easy

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.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]

• 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]

• 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.

Follow up:

• Could you find an O(nums1.length + nums2.length) solution?

Solution

class Solution:
    # Iteration
    # Time Complexity: BigO(N^2)
    # Space Complexity: BigO(N)
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        stack = []
        dict = {}
        res = []

        for i in nums2:
            while len(stack) > 0 and i > stack[-1]:
                dict[stack.pop()] = i
            stack.append(i)

        for i in nums1:
            res.append(dict.get(i, -1))

        return res

/**
 * Iteration
 * Time Complexity: BigO(N^2)
 * Space Complexity: BigO(N)
 */
function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
  let res: number[] = [];
  nums1.forEach((num1) => {
    let idx = nums2.indexOf(num1);
    for (let j = idx + 1; j < nums2.length; j++) {
      if (nums2[j] > num1) {
        res.push(nums2[j]);
        return;
      }
    }
    res.push(-1);
  });
  return res;
}