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 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

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.