Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/496.html

496. Next Greater Element I (Easy)

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are 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, output -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.


Note:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

Companies:

Related Topics:
Stack

Similar Questions:

Solution 1. Monotonic Stack

// OJ: https://leetcode.com/problems/next-greater-element-i/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
stack<int> s;
unordered_map<int, int> m;
for (int n : nums) {
while (s.size() && s.top() < n) {
m[s.top()] = n;
s.pop();
}
s.push(n);
}
vector<int> ans;
for (int n : findNums) ans.push_back(m.count(n) ? m[n] : -1);
return ans;
}
};

• class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
int length1 = nums1.length, length2 = nums2.length;
List<Integer> list2 = new ArrayList<Integer>();
for (int i = 0; i < length2; i++)
int[] nextGreaterElements1 = new int[length1];
int[] nextGreaterElements2 = new int[length2];
Stack<Integer> stack = new Stack<Integer>();
for (int i = 0; i < length2; i++) {
int num = nums2[i];
while (!stack.isEmpty() && stack.peek() < num) {
int prevNum = stack.pop();
int index = list2.indexOf(prevNum);
nextGreaterElements2[index] = num;
}
stack.push(num);
}
while (!stack.isEmpty()) {
int num = stack.pop();
int index = list2.indexOf(num);
nextGreaterElements2[index] = -1;
}
for (int i = 0; i < length1; i++) {
int num = nums1[i];
int index = list2.indexOf(num);
nextGreaterElements1[i] = nextGreaterElements2[index];
}
return nextGreaterElements1;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/next-greater-element-i/
// Time: O(A + B)
// Space: O(B) extra space
class Solution {
public:
vector<int> nextGreaterElement(vector<int>& A, vector<int>& B) {
unordered_map<int, int> m;
stack<int> s;
for (int n : B) {
while (s.size() && n > s.top()) {
m[s.top()] = n;
s.pop();
}
s.push(n);
}
vector<int> ans;
for (int n : A) ans.push_back(m.count(n) ? m[n] : -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]

############

class Solution(object):
def nextGreaterElement(self, findNums, nums):
"""
:type findNums: List[int]
:type nums: List[int]
:rtype: List[int]
"""
d = {}
ans = [-1] * len(findNums)
for i, num in enumerate(findNums):
d[num] = i
stack = []
for num in nums:
while stack and stack[-1] < num:
top = stack.pop()
if top in d:
ans[d[top]] = num
stack.append(num)
return ans


• 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>>()
}
}