LeetCode-496. Next Greater Element I [C++][Java]

LeetCode-496. Next Greater Element Ihttps://leetcode.com/problems/next-greater-element-i/

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] <= 10^4
• All integers in nums1 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

class Solution {
public:
    vector nextGreaterElement(vector& nums1, vector& nums2) {
        int m = nums1.size();
        int n = nums2.size();
        vector res(m);
        for (int i = 0; i < m; ++i) {
            int j = 0;
            while (j < n && nums2[j] != nums1[i]) {++j;}
            int k = j + 1;
            while (k < n && nums2[k] < nums2[j]) {++k;}
            res[i] = k < n ? nums2[k] : -1;
        }
        return res;
    }
};

class Solution {
public:
    vector nextGreaterElement(vector& nums1, vector& nums2) {
        stack st;
        int n = nums2.size();
        map v;
        for (int i = n-1; i>=0; i--) {
            if (st.empty()) {
                v[nums2[i]] = -1;
                st.push(nums2[i]);
            } else {
                while (!st.empty() && st.top() <= nums2[i]) {st.pop();}
                if (st.empty()) {v[nums2[i]] = -1;}
                else {v[nums2[i]] = st.top();}
                st.push(nums2[i]);
            }
        }
        n = nums1.size();
        vector ans(n);
        for (int i=0;i

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int n1 = nums1.length;
        int n2 = nums2.length;
        int[] ans = new int[n1];
        // To hold the numbers in the right which can be greater than current number
        Stack stack = new Stack<>();
        // To store the mapping of number with it's greater number in right
        Map greaterInRight = new HashMap<>();
        // Iterate from last number to the first number
        for(int i=n2-1; i>=0; i--) {
            // Remove numbers from top of the stack until a greater number is found
            while( !stack.isEmpty() && stack.peek() < nums2[i] ) {
                stack.pop();
            }
            /* If a stack is not empty, then the top of the stack contains the greater
            number than current number, else there is no number in right which is 
            greater than current number */
            greaterInRight.put(nums2[i], stack.isEmpty() ? -1 : stack.peek() );
            /* Push current number into stack, as this number can be greater number
            for the numbers in left */
            stack.push( nums2[i] );
        }
        // Use the map to get the number in right greater than each number in {nums1}
        for(int i=0; i