Old fashioned boolean

People don't give much credit anymore to George Boole (George Boole - Wikipedia), but that's a mistake, a simple boolean operation can save you a lot of time and space.

This problem exemplifies it. Given the low boundaries (N=50), an N^2 solution works fine and the use of a simple boolean array makes it an elegant solution too. Code is down below, cheers, ACC.

Find the Prefix Common Array of Two Arrays - LeetCode

2657. Find the Prefix Common Array of Two Arrays

Medium

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.

public int[] FindThePrefixCommonArray(int[] A, int[] B)
{
    int[] C = new int[A.Length];

    bool[] inA = new bool[A.Length + 1];
    bool[] inB = new bool[B.Length + 1];

    for (int i = 0; i < C.Length; i++)
    {
        inA[A[i]] = true;
        inB[B[i]] = true;

        C[i] = 0;
        for (int j = 0; j < inA.Length; j++)
            C[i] = (inA[j] & inB[j]) ? C[i] + 1 : C[i];
    }

    return C;
}