最长公共子序列转化为最长上升子序列

leetcode https://leetcode.com/problems/minimum-operations-to-make-a-subsequence/

Weekly Contest 222

1713. Minimum Operations to Make a Subsequence

You are given an array target that consists of distinct integers and another integer array arr that can have duplicates.

In one operation, you can insert any integer at any position in arr. For example, if arr = [1,4,1,2], you can add 3 in the middle and make it [1,4,3,1,2]. Note that you can insert the integer at the very beginning or end of the array.

Return the minimum number of operations needed to make target a subsequence of arr.

A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

 

Example 1:

Input: target = [5,1,3], arr = [9,4,2,3,4]
Output: 2
Explanation: You can add 5 and 1 in such a way that makes arr = [5,9,4,1,2,3,4], 
then target will be a subsequence of arr.

Constraints:

• 1 <= target.length, arr.length <= 105
• 1 <= target[i], arr[i] <= 109
• target contains no duplicates.

 

 

    public int minOperations(int[] target , int[] arr) {
        Map map = new HashMap();
        int n = target.length, m = arr.length;
        for(int i = 0; i < n; i++){
            map.put(target[i], i + 1);
        }
        int[] arr2 = new int[m + 1];
        for(int i = 0; i < m; i++){
            if(map.containsKey(arr[i])){
                arr2[i + 1] = map.get(arr[i]);
            }
        }

        List list = new ArrayList();
        for(int i = 1; i <= m; i++){
            if(arr2[i] == 0) continue;
            if(list.isEmpty() || arr2[i] > list.get(list.size() - 1)){
                list.add(arr2[i]);
            }else{
                int l = 0, r = list.size() - 1;
                while(l < r){
                    int mid = l + r >> 1;
                    if(list.get(mid) >= arr2[i]) r = mid;
                    else l = mid + 1;
                }
                list.set(l, arr2[i]);
            }
        }
        return n - list.size();
    }