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2657. Find the Prefix Common Array of Two Arrays
Description
You are given two 0indexed 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.
Solutions
Solution 1: Counting
We can use two arrays $cnt1$ and $cnt2$ to record the occurrence times of each element in arrays $A$ and $B$ respectively, and use an array $ans$ to record the answer.
Traverse arrays $A$ and $B$, increment the occurrence times of $A[i]$ in $cnt1$, and increment the occurrence times of $B[i]$ in $cnt2$. Then enumerate $j \in [1,n]$, calculate the minimum occurrence times of each element $j$ in $cnt1$ and $cnt2$, and accumulate them into $ans[i]$.
After the traversal, return the answer array $ans$.
The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of arrays $A$ and $B$.
Solution 2: Bit Operation (XOR Operation)
We can use an array $vis$ of length $n+1$ to record the occurrence situation of each element in arrays $A$ and $B$, the initial value of array $vis$ is $1$. In addition, we use a variable $s$ to record the current number of common elements.
Next, we traverse arrays $A$ and $B$, update $vis[A[i]] = vis[A[i]] \oplus 1$, and update $vis[B[i]] = vis[B[i]] \oplus 1$, where $\oplus$ represents XOR operation.
If at the current position, the element $A[i]$ has appeared twice (i.e., it has appeared in both arrays $A$ and $B$), then the value of $vis[A[i]]$ will be $1$, and we increment $s$. Similarly, if the element $B[i]$ has appeared twice, then the value of $vis[B[i]]$ will be $1$, and we increment $s$. Then add the value of $s$ to the answer array $ans$.
After the traversal, return the answer array $ans$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of arrays $A$ and $B$.

class Solution { public int[] findThePrefixCommonArray(int[] A, int[] B) { int n = A.length; int[] ans = new int[n]; int[] cnt1 = new int[n + 1]; int[] cnt2 = new int[n + 1]; for (int i = 0; i < n; ++i) { ++cnt1[A[i]]; ++cnt2[B[i]]; for (int j = 1; j <= n; ++j) { ans[i] += Math.min(cnt1[j], cnt2[j]); } } return ans; } }

class Solution { public: vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) { int n = A.size(); vector<int> ans(n); vector<int> cnt1(n + 1), cnt2(n + 1); for (int i = 0; i < n; ++i) { ++cnt1[A[i]]; ++cnt2[B[i]]; for (int j = 1; j <= n; ++j) { ans[i] += min(cnt1[j], cnt2[j]); } } return ans; } };

class Solution: def findThePrefixCommonArray(self, A: List[int], B: List[int]) > List[int]: ans = [] cnt1 = Counter() cnt2 = Counter() for a, b in zip(A, B): cnt1[a] += 1 cnt2[b] += 1 t = sum(min(v, cnt2[x]) for x, v in cnt1.items()) ans.append(t) return ans

func findThePrefixCommonArray(A []int, B []int) []int { n := len(A) cnt1 := make([]int, n+1) cnt2 := make([]int, n+1) ans := make([]int, n) for i, a := range A { b := B[i] cnt1[a]++ cnt2[b]++ for j := 1; j <= n; j++ { ans[i] += min(cnt1[j], cnt2[j]) } } return ans }

function findThePrefixCommonArray(A: number[], B: number[]): number[] { const n = A.length; const cnt1: number[] = Array(n + 1).fill(0); const cnt2: number[] = Array(n + 1).fill(0); const ans: number[] = Array(n).fill(0); for (let i = 0; i < n; ++i) { ++cnt1[A[i]]; ++cnt2[B[i]]; for (let j = 1; j <= n; ++j) { ans[i] += Math.min(cnt1[j], cnt2[j]); } } return ans; }