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.

Example 2:

Input: target = [6,4,8,1,3,2], arr = [4,7,6,2,3,8,6,1]
Output: 3

Constraints:

• 1 <= target.length, arr.length <= 10^5
• 1 <= target[i], arr[i] <= 10^9
• target contains no duplicates.

We start by observing the follwing key points:

1.  Numbers that do not exist in target array do not matter, they can be ignored;

2. Target array only has distinct integers, this means we can map each unique integer to its idx in target array;

3. For array arr, if we ignore numbers that do not exist in target but replace all the existing numbers with their idx in target, this problem becomes this: Given N unique increasing numbers from 0 to N - 1, (N is target array's length) and an array A that can only contains number from 0 to N - 1, find the minimum insertions needed such that A has an increasing subsequence from 0 to N - 1.

We can further reduce the above problem to: find the longest increasing subsequence in A, call it Length(LIS), N - Length(LIS) will be the minimum insertions needed.

class Solution {
    public int minOperations(int[] target, int[] arr) {
        Map<Integer, Integer> map = new HashMap<>();
        int n = target.length;
        for(int i = 0; i < n; i++) {
            map.put(target[i], i);
        }
        List<Integer> list = new ArrayList<>();
        for(int v : arr) {
            if(map.containsKey(v)) {
                list.add(map.get(v));
            }
        }
        return n - lengthOfLIS(list);
    }
    private int lengthOfLIS(List<Integer> list) {
        List<ArrayDeque<Integer>> qList = new ArrayList<>();
        for(int v : list) {
            int idx = binarySearch(qList, v);
            if(idx == qList.size()) {
                qList.add(new ArrayDeque<>());
            }
            ArrayDeque<Integer> q = qList.get(idx);
            q.addFirst(v);
        }
        return qList.size();
    }
    private int binarySearch(List<ArrayDeque<Integer>> qList, int v) {
        if(qList.size() > 0) {
            int l = 0, r = qList.size() - 1;
            while(l < r) {
                int mid = l + (r - l) / 2;
                if(qList.get(mid).peekFirst() < v) {
                    l = mid + 1;
                }
                else {
                    r = mid;
                }
            }
            if(qList.get(l).peekFirst() >= v) {
                return l;
            }
        }
        return qList.size();
    }
}

Related Problems

[LeetCode 300] Longest Increasing Subsequence

标签：1713,Operations,arr,target,int,Make,subsequence,array,qList
来源： https://www.cnblogs.com/lz87/p/14245541.html