Introduction

In this blog post, we will explore how to solve the LeetCode problem "2657" 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 two 0-indexed integer permutations A and B of length n. A prefix common array of A and B is an array C such that C[i] is equal to the count of numbers that are present at or before the index i in both A and B. Return the prefix common array of A and B. A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once. Example 1: Input: A = [1,3,2,4], B = [3,1,2,4] Output: [0,2,3,4] Explanation: At i = 0: no number is common, so C[0] = 0. At i = 1: 1 and 3 are common in A and B, so C[1] = 2. At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3. At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4. Example 2: Input: A = [2,3,1], B = [3,1,2] Output: [0,1,3] Explanation: At i = 0: no number is common, so C[0] = 0. At i = 1: only 3 is common in A and B, so C[1] = 1. At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3. Constraints: 1 <= A.length == B.length == n <= 50 1 <= A[i], B[i] <= n It is guaranteed that A and B are both a permutation of n integers.

Explanation

Here's the breakdown of the solution:

• Efficient Common Element Tracking: Use a boolean array (or set) to efficiently track the presence of numbers encountered in both permutations A and B.
• Incremental Common Count: Iterate through the permutations simultaneously, updating the common count as new common elements are found.

• Prefix Array Construction: Build the prefix common array by storing the cumulative common counts at each index.

• Runtime Complexity: O(n), where n is the length of the input arrays. Storage Complexity: O(n) due to the boolean array.

Code

    def prefix_common_array(A, B):
    n = len(A)
    C = [0] * n
    seen = [False] * (n + 1)  # Use boolean array instead of set for slight potential performance gain (constant factor)

    count = 0
    for i in range(n):
        if not seen[A[i]]:
            if A[i] in B[:i+1]: # Optimization: Directly check if element is in the prefix of B. More efficient for small constraint n <= 50
                count += 1
                seen[A[i]] = True
        if not seen[B[i]]:
            if B[i] in A[:i+1]: # Optimization: Directly check if element is in the prefix of A
                count += 1
                seen[B[i]] = True
        C[i] = count
    return C