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Home Code Solutions

Next Greater Element I – LeetCode Solution

Next Greater Element I - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 10, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

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  • Next Greater Element I – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Next Greater Element I LeetCode Solution
  • Java Next Greater Element I LeetCode Solution
  • Python 3 Next Greater Element I LeetCode Solution
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Next Greater Element I – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

C++ Next Greater Element I LeetCode Solution


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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;
    }
};

Java Next Greater Element I LeetCode Solution


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 public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
        Stack<Integer> stack = new Stack<>();
        for (int num : nums) {
            while (!stack.isEmpty() && stack.peek() < num)
                map.put(stack.pop(), num);
            stack.push(num);
        }   
        for (int i = 0; i < findNums.length; i++)
            findNums[i] = map.getOrDefault(findNums[i], -1);
        return findNums;
    }

 



Python 3 Next Greater Element I LeetCode Solution


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def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
	if not nums2:
		return None

	mapping = {}
	result = []
	stack = []
	stack.append(nums2[0])

	for i in range(1, len(nums2)):
		while stack and nums2[i] > stack[-1]:       # if stack is not empty, then compare it's last element with nums2[i]
			mapping[stack[-1]] = nums2[i]           # if the new element is greater than stack's top element, then add this to dictionary 
			stack.pop()                             # since we found a pair for the top element, remove it.
		stack.append(nums2[i])                      # add the element nums2[i] to the stack because we need to find a number greater than this

	for element in stack:                           # if there are elements in the stack for which we didn't find a greater number, map them to -1
		mapping[element] = -1

	for i in range(len(nums1)):
		result.append(mapping[nums1[i]])
	return result



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264


Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195


Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140


Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96


Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66


Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40


Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31


Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21


Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16


Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9


Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7


Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4


Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3


Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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