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Solution: Minimum Operations to Make a Subsequence

seanpgallivan
Fledgling software developer; the struggle is a Rational Approximation.
・3 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.


Leetcode Problem #1713 (Hard): Minimum Operations to Make a Subsequence


Description:

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], while [2,4,2] is not.


Examples:

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

Idea:

Normally, we'd solve this problem by identifying the longest common subsequence, as that would also indicate how many elements would therefore need to be inserted to make the target array (T) a possible match. LCS algorithms have an O(m * n) time complexity, however, which is far too long in this case.

This solution is actually much more straightforward once we realize that T has distinct elements. That means that instead of an LCS approach, we can instead treat the elements of T as their index and solve this using a largest increasing subsequence approach instead, with a time complexity of O(n * log n).

In a LIS solution, we first need to create an index map (imap) to use a reference. Since we only need the length of the larest subsequence rather than needing to be able to reconstruct it, we just need to use an array (lis) where lis[i] will keep track of the last number in the most efficient (i-1)-length subsequence.

In other words, lis[4] would be the last number in the lexicographically smallest three-number subsequence. Because each of these subsequences must be increasing by definition, and because each entry in lis represents the best possible version of each length of subsequence, then lis is by its very nature an ordered array.

This means that any new number we come across while iterating through A (and referencing A[i] through imap) can be used to replace the first available entry of lis that is larger. And since lis is ordered, we can use a simple binary search to find the appropriate replacement index of lis.

Once we're done iterating through A, the length of the longest increasing susbequence will be equal to the length of lis, which is likewise then the length of the longest common subsequence between T and A. All we need to do at that point is return its difference from T's length to find out how many operations it would take to complete T.


Javascript Code:

var minOperations = function(T, A) {
    let imap = new Map(), lis = []
    for (let i = 0; i < T.length; i++) imap.set(T[i], i)
    for (let i = 0; i < A.length; i++) {
        let index = imap.get(A[i])
        if (index !== undefined)
            lis[find(index, lis)] = index
    }
    return T.length - lis.length
};

const find = (target, arr, left=0, right=arr.length) => {
    while (left <= right) {
        let mid = left + right >> 1
        if (arr[mid] < target) left = mid + 1
        else right = mid - 1
    }
    return left
}
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