Next Greater Element (DSA)

 

Problem Statement:(👈click here for GFG)

Given an array arr[ ] of size N having elements, the task is to find the next greater element for each element of the array in order of their appearance in the array.
Next greater element of an element in the array is the nearest element on the right which is greater than the current element.
If there does not exist next greater of current element, then next greater element for current element is -1. For example, next greater of the last element is always -1.

Example 1:

Input: 
N = 4, arr[] = [1 3 2 4]
Output:
3 4 4 -1
Explanation:
In the array, the next larger element 
to 1 is 3 , 3 is 4 , 2 is 4 and for 4 ? 
since it doesn't exist, it is -1.

Example 2:

Input: 
N = 5, arr[] [6 8 0 1 3]
Output:
8 -1 1 3 -1
Explanation:
In the array, the next larger element to 
6 is 8, for 8 there is no larger elements 
hence it is -1, for 0 it is 1 , for 1 it 
is 3 and then for 3 there is no larger 
element on right and hence -1.

Your Task:
This is a function problem. You only need to complete the function nextLargerElement() that takes list of integers arr[ ] and N as input parameters and returns list of integers of length N denoting the next greater elements for all the corresponding elements in the input array.

Expected Time Complexity : O(N)
Expected Auxiliary Space : O(N)

Constraints:
1 ≤ N ≤ 106
0 ≤ Ai ≤ 1018

Solution:

class Solution {
    //Function to find the next greater element for each element of the array.
    nextLargerElement(arr, n) {
        let stack = [];
        let result = [];
        
        // traverse the array from right to left
        for (let i = n - 1; i >= 0; i--) {
            // pop all the elements from the stack which are smaller than or equal to the current element
            while (stack.length > 0 && stack[stack.length - 1] <= arr[i]) {
                stack.pop();
            }
            
            // next greater element for the current element is the top element of the stack (if not empty)
            if (stack.length > 0) {
                result[i] = stack[stack.length - 1];
            } else {
                result[i] = -1;
            }
            
            // push the current element to the stack
            stack.push(arr[i]);
        }
        
        return result;
    }
}