In this post, we are going to solve the 496. Next Greater Element I problem of Leetcode. This problem 496. Next Greater Element I is a Leetcode easy level problem. Let’s see code, 496. Next Greater Element I.
Problem
In this Leetcode Next Greater Element I problem solution,
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
andnums2
are unique. - All the integers of
nums1
also appear innums2
.
Now, let’s see the code of 496. Next Greater Element I – Leetcode Solution.
Next Greater Element I – Leetcode Solution
496. Next Greater Element I – Solution in Java
class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { int[] ans = new int[nums1.length]; for(int i=0;i<nums1.length;i++){ ans[i] = nextGreaterElement(nums2,nums1[i]); } return ans; } public int nextGreaterElement(int[] nums, int tar){ Stack<Integer> s = new Stack<>(); for(int e : nums){ s.add(e); } int ans = -1; while(!(s.isEmpty())){ int top = s.peek(); if(top == tar) return ans; s.pop(); if(top > tar){ ans = top; } } return ans; } }
496. Next Greater Element I – Solution in C++
class Solution { public: vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) { vector<int> ans(nums1.size()); for(int i=0;i<nums1.size();i++){ ans[i] = nextGreaterElement(nums2,nums1[i]); } return ans; } int nextGreaterElement(vector<int>& nums, int tar){ stack<int> s; for(int e : nums){ s.push(e); } int ans = -1; while((s.size()>0)){ int top = s.top(); if(top == tar) return ans; s.pop(); if(top > tar){ ans = top; } } return ans; } };
496. Next Greater Element I – Solution in Python
This is a python 3 solution of Leetcode 496. Next Greater Element I.
class Solution: def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: ans = {} res = [] stack = [] for n2 in nums2: while stack and n2 > stack[-1]: ans[stack.pop()] = n2 stack.append(n2) for n1 in nums1: res.append(ans.get(n1, -1)) return res
Note: This problem 496. Next Greater Element I is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.