503. Next Greater Element II

Problem Statement

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return -1 for this number.

Example 1:

Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number. 
The second 1's next greater number needs to search circularly, which is also 2.

Example 2:

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

Constraints:

1 <= nums.length <= 104
-109 <= nums[i] <= 109

Intuition

Approach:
Array has duplicates, So Store index in map rather than actual values

Links

https://leetcode.com/problems/next-greater-element-ii/description/

Video Links

Approach 1:

C++

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        int n=nums.size();
        unordered_map<int, int>mp;
        stack<int> st;
        for(int i=0; i<2*n; i++) {
            while(!st.empty() and nums[st.top()] < nums[i%n]){
                mp[st.top()] = nums[i%n];
                st.pop();
            }
            st.push(i%n);
        }

        vector<int> ans;
        for(int i=0; i<n; i++){
            if(mp.find(i) != mp.end())
                ans.push_back(mp[i]);
            else
                ans.push_back(-1);
        }
            

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++