496. Next Greater Element I

MA:
class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int[] nums3=new int[nums1.length];
        for(int i=0;i<nums1.length;i++){
            int max=-1;
            nums3[i]=max;
            int target=nums1[i];
            boolean findTarget=false;
            for(int j=0;j<nums2.length;j++){
                if(nums2[j]==nums1[i] && findTarget==false){
                    findTarget=true;
                    continue;
                }
                if(nums2[j]>target&&findTarget==true){
                    max=nums2[j];
                    break;
                }
            }
            findTarget=false;;
            nums3[i]=max;
        }     
        return nums3;
    }
}

OA:
 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;
    }

Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence



We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all the popped ones, their next greater element is x
For example [9, 8, 7, 3, 2, 1, 6]
The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6

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