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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:
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
Companies:
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Related Topics:
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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;
}
};
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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++) list2.add(nums2[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; } }
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// 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; } };
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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
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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 }
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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); }
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/** * @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); };
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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>>() } }