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

Solution 1: Monotonic Stack

We can traverse the array $\textit{nums2}$ from right to left, maintaining a stack $\textit{stk}$ that is monotonically increasing from top to bottom. We use a hash table $\textit{d}$ to record the next greater element for each element.

When we encounter an element $x$, if the stack is not empty and the top element of the stack is less than $x$, we keep popping the top elements until the stack is empty or the top element is greater than or equal to $x$. At this point, if the stack is not empty, the top element of the stack is the next greater element for $x$. Otherwise, $x$ has no next greater element.

Finally, we traverse the array $\textit{nums1}$ and use the hash table $\textit{d}$ to get the answer.

The time complexity is $O(m + n)$, and the space complexity is $O(n)$. Here, $m$ and $n$ are the lengths of the arrays $\textit{nums1}$ and $\textit{nums2}$, respectively.

Python3JavaC++GoTypeScriptRustJavaScript

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

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Deque<Integer> stk = new ArrayDeque<>();
        int m = nums1.length, n = nums2.length;
        Map<Integer, Integer> d = new HashMap(n);
        for (int i = n - 1; i >= 0; --i) {
            int x = nums2[i];
            while (!stk.isEmpty() && stk.peek() < x) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                d.put(x, stk.peek());
            }
            stk.push(x);
        }
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = d.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> d;
        ranges::reverse(nums2);
        for (int x : nums2) {
            while (!stk.empty() && stk.top() < x) {
                stk.pop();
            }
            if (!stk.empty()) {
                d[x] = stk.top();
            }
            stk.push(x);
        }
        vector<int> ans;
        for (int x : nums1) {
            ans.push_back(d.contains(x) ? d[x] : -1);
        }
        return ans;
    }
};

func nextGreaterElement(nums1 []int, nums2 []int) (ans []int) {
    stk := []int{}
    d := map[int]int{}
    for i := len(nums2) - 1; i >= 0; i-- {
        x := nums2[i]
        for len(stk) > 0 && stk[len(stk)-1] < x {
            stk = stk[:len(stk)-1]
        }
        if len(stk) > 0 {
            d[x] = stk[len(stk)-1]
        }
        stk = append(stk, x)
    }
    for _, x := range nums1 {
        if v, ok := d[x]; ok {
            ans = append(ans, v)
        } else {
            ans = append(ans, -1)
        }
    }
    return
}

function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
    const stk: number[] = [];
    const d: Record<number, number> = {};
    for (const x of nums2.reverse()) {
        while (stk.length && stk.at(-1)! < x) {
            stk.pop();
        }
        d[x] = stk.length ? stk.at(-1)! : -1;
        stk.push(x);
    }
    return nums1.map(x => d[x]);
}

use std::collections::HashMap;

impl Solution {
    pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let mut stk = Vec::new();
        let mut d = HashMap::new();
        for &x in nums2.iter().rev() {
            while let Some(&top) = stk.last() {
                if top <= x {
                    stk.pop();
                } else {
                    break;
                }
            }
            if let Some(&top) = stk.last() {
                d.insert(x, top);
            }
            stk.push(x);
        }

        nums1
            .into_iter()
            .map(|x| *d.get(&x).unwrap_or(&-1))
            .collect()
    }
}

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var nextGreaterElement = function (nums1, nums2) {
    const stk = [];
    const d = {};
    for (const x of nums2.reverse()) {
        while (stk.length && stk.at(-1) < x) {
            stk.pop();
        }
        d[x] = stk.length ? stk.at(-1) : -1;
        stk.push(x);
    }
    return nums1.map(x => d[x]);
};

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