LeetCode in Kotlin
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 <= 501 <= A[i], B[i] <= nIt is guaranteed that A and B are both a permutation of n integers.
Solution
class Solution {
fun findThePrefixCommonArray(a: IntArray, b: IntArray): IntArray {
val hsA = HashSet<Int>()
val hsB = HashSet<Int>()
val addedA = HashSet<Int>()
val addedB = HashSet<Int>()
val res = IntArray(a.size)
for (i in a.indices) {
val numA = a[i]
val numB = b[i]
hsA.add(numA)
hsB.add(numB)
if (i > 0) res[i] += res[i - 1] else res[i] = 0
if (numA in hsB && numA !in addedB) {
addedA.add(numA)
res[i] += 1
}
if (numB in hsA && numB !in addedA) {
addedB.add(numB)
res[i] += 1
}
}
return res
}
}