README_EN.md

496. Next Greater Element I

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Description

You are given two integer arrays nums1 and nums2 both of unique elements, where nums1 is a subset of nums2.

Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, return -1 for this number.

 

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.
For number 4 in the first array, there is no next greater number for it in the second array, so output -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.

 

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solutions

Python3

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        mp = {}
        stk = []
        for num in nums2:
            while stk and stk[-1] < num:
                mp[stk.pop()] = num
            stk.append(num)
        return [mp.get(num, -1) for num in nums1]

Java

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Deque<Integer> stk = new ArrayDeque<>();
        Map<Integer, Integer> mp = new HashMap<>();
        for (int num : nums2) {
            while (!stk.isEmpty() && stk.peek() < num) {
                mp.put(stk.pop(), num);
            }
            stk.push(num);
        }
        int n = nums1.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = mp.getOrDefault(nums1[i], -1);
        }
        return ans;
    }
}

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var nextGreaterElement = function(nums1, nums2) {
    let stk = [];
    let nextGreater = {};
    for (let num of nums2) {
        while (stk && stk[stk.length - 1] < num) {
            nextGreater[stk.pop()] = num;
        }
        stk.push(num);
    }
    return nums1.map(e => nextGreater[e] || -1);
};

C++

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        stack<int> stk;
        unordered_map<int, int> mp;
        for (int num : nums2)
        {
            while (!stk.empty() && stk.top() < num)
            {
                mp[stk.top()] = num;
                stk.pop();
            }
            stk.push(num);
        }
        vector<int> ans;
        for (int num : nums1) ans.push_back(mp.count(num) ? mp[num] : -1);
        return ans;
    }
};

Go

func nextGreaterElement(nums1 []int, nums2 []int) []int {
	var stk []int
	mp := make(map[int]int)
	for _, num := range nums2 {
		for len(stk) > 0 && stk[len(stk)-1] < num {
			mp[stk[len(stk)-1]] = num
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, num)
	}
	var ans []int
	for _, num := range nums1 {
		val, ok := mp[num]
		if !ok {
			val = -1
		}
		ans = append(ans, val)
	}
	return ans
}

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