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496. Next Greater Element I

Description

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.

 

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

Solutions

  • 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;
    }
}

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

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


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

  • function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
    const map = new Map<number, number>();
    const stack: number[] = [Infinity];
    for (const num of nums2) {
        while (num > stack[stack.length - 1]) {
            map.set(stack.pop(), num);
        }
        stack.push(num);
    }
    return nums1.map(num => map.get(num) || -1);
}


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


  • use std::collections::HashMap;

impl Solution {
    pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let mut map = HashMap::new();
        let mut stack = Vec::new();
        for num in nums2 {
            while num > *stack.last().unwrap_or(&i32::MAX) {
                map.insert(stack.pop().unwrap(), num);
            }
            stack.push(num);
        }
        nums1
            .iter()
            .map(|n| *map.get(n).unwrap_or(&-1))
            .collect::<Vec<i32>>()
    }
}

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