Introduction

In this blog post, we will explore how to solve the LeetCode problem "1713" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a breakdown of the approach and the Python code:

• Key Idea: The problem is equivalent to finding the longest common subsequence (LCS) of target and arr. The minimum number of insertions needed would then be len(target) - len(LCS). We can't directly compute LCS due to the constraints. Since target contains distinct integers, we can use a hashmap to store the index of each number in target.
• Transform to LIS: We iterate through arr, and if an element exists in target, we replace the element in arr with its index in target. Then, the problem becomes finding the Longest Increasing Subsequence (LIS) within the transformed arr. The length of the LIS is the length of the LCS.

• LIS Efficiently: We use a standard efficient algorithm for LIS based on binary search.

• Complexity: Time Complexity: O(N log K), where N is the length of arr and K is the length of target. Space Complexity: O(K), due to the dp array and the hashmap.

Code

    def minOperations(target, arr):
    """
    Calculates the minimum number of operations (insertions) needed to make target a subsequence of arr.

    Args:
        target (List[int]): The target array containing distinct integers.
        arr (List[int]): The array to modify by inserting elements.

    Returns:
        int: The minimum number of operations needed.
    """

    target_map = {val: i for i, val in enumerate(target)}
    nums = []
    for val in arr:
        if val in target_map:
            nums.append(target_map[val])

    if not nums:
        return len(target)

    dp = []
    for num in nums:
        if not dp or num > dp[-1]:
            dp.append(num)
        else:
            l, r = 0, len(dp) - 1
            while l <= r:
                mid = (l + r) // 2
                if dp[mid] < num:
                    l = mid + 1
                else:
                    r = mid - 1
            dp[l] = num

    return len(target) - len(dp)