2657. Find the Prefix Common Array of Two Arrays

Description

You are given two 0-indexed integerpermutations 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.

 

Solutions

Solution: Hash Map

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} A
 * @param {number[]} B
 * @return {number[]}
 */
const findThePrefixCommonArray = function (A, B) {
  const n = A.length;
  const counts = Array.from({ length: n + 1 }, () => 0);
  const result = [];
  let prefixCommon = 0;

  for (let index = 0; index < n; index++) {
    const a = A[index];
    const b = B[index];

    counts[a] += 1;
    if (counts[a] === 2) prefixCommon += 1;

    counts[b] += 1;
    if (counts[b] === 2) prefixCommon += 1;

    result.push(prefixCommon);
  }
  return result;
};