496.next-greater-element-i.cpp

/*
 * @lc app=leetcode id=496 lang=cpp
 *
 * [496] Next Greater Element I
 *
 * https://leetcode.com/problems/next-greater-element-i/description/
 *
 * algorithms
 * Easy (57.75%)
 * Total Accepted:    77K
 * Total Submissions: 133.2K
 * Testcase Example:  '[4,1,2]\n[1,3,4,2]'
 *
 *
 * 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 and nums2 are unique.
 * The length of both nums1 and nums2 would not exceed 1000.
 *
 *
 */
class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        unordered_map<int, int> mm;
        stack<int> stk;
        for(int i=(int)nums.size()-1;i>=0;i--){
          int e = nums[i];
          while(!stk.empty() && stk.top() < e)
            stk.pop();
          if(stk.empty())
              mm[e] = -1;
          else
              mm[e] = stk.top();
          stk.push(e);
          // cout<<stk.top()<<endl;
        }
        vector<int> ans;
        for(auto e: findNums)
          ans.push_back(mm[e]);
        return ans;
    }
};