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

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

 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

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
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Two Pointers-151
Bit Manipulation-140

Stack-133
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Design-116
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Prefix Sum-96

Backtracking-92
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Sliding Window-73
Linked List-69
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Ordered Set-48
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Recursion-43
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Enumeration-39
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Queue-33
Memoization-32
Topological Sort-31

Geometry-30
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Game Theory-24
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Binary Indexed Tree-21

Interactive-18
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Concurrency-9
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Bucket Sort-6
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Shell-4

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Strongly Connected Componen-t2
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