Next Greater Element

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.

Solution : Naive

We can easily solve this problme by using two loop. But the time complexity for this solution is O(n^2).

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
    vector<int> res;
    for(int i=0; i<findNums.size(); i++) {
        int temp = -1;
        for(int j=nums.size()-1; j >= 0; j--) {
            if(nums[j] == findNums[i]) break;
            if(nums[j] > findNums[i]) temp = nums[j];
        }
        res.push_back(temp);
    }
    return res;
}

Solution: Using stack

First traverse the array to find the next greater element for all items, and store them in a hash table. Traverse the subset array, using the hash table to find the next greater element in O(1).

Note: The key is to use stack to find the next greater element in one traversal.

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
    stack<int> todo;
    unordered_map<int, int> m;
    for (int i: nums) {
        while (!todo.empty() && i > todo.top()) {
            m[todo.top()] = i;
            todo.pop();
        }
        todo.push(i);
    }
    vector<int> res;
    for (int i: findNums) {
        auto it = m.find(i);
        res.push_back(it == m.end() ? -1 : it->second);
    }
    return res;
}