Question

2657. Find the Prefix Common Array of Two Arrays

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Description

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.

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 $B$.

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

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;
  }
};

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;
}

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:
  def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
    ans = []
    vis = [1] * (len(A) + 1)
    s = 0
    for a, b in zip(A, B):
      vis[a] ^= 1
      s += vis[a]
      vis[b] ^= 1
      s += vis[b]
      ans.append(s)
    return ans

class Solution {
  public int[] findThePrefixCommonArray(int[] A, int[] B) {
    int n = A.length;
    int[] ans = new int[n];
    int[] vis = new int[n + 1];
    Arrays.fill(vis, 1);
    int s = 0;
    for (int i = 0; i < n; ++i) {
      vis[A[i]] ^= 1;
      s += vis[A[i]];
      vis[B[i]] ^= 1;
      s += vis[B[i]];
      ans[i] = s;
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
    int n = A.size();
    vector<int> ans;
    vector<int> vis(n + 1, 1);
    int s = 0;
    for (int i = 0; i < n; ++i) {
      vis[A[i]] ^= 1;
      s += vis[A[i]];
      vis[B[i]] ^= 1;
      s += vis[B[i]];
      ans.push_back(s);
    }
    return ans;
  }
};

func findThePrefixCommonArray(A []int, B []int) (ans []int) {
  vis := make([]int, len(A)+1)
  for i := range vis {
    vis[i] = 1
  }
  s := 0
  for i, a := range A {
    b := B[i]
    vis[a] ^= 1
    s += vis[a]
    vis[b] ^= 1
    s += vis[b]
    ans = append(ans, s)
  }
  return
}

function findThePrefixCommonArray(A: number[], B: number[]): number[] {
  const n = A.length;
  const vis: number[] = Array(n + 1).fill(1);
  const ans: number[] = [];
  let s = 0;
  for (let i = 0; i < n; ++i) {
    const [a, b] = [A[i], B[i]];
    vis[a] ^= 1;
    s += vis[a];
    vis[b] ^= 1;
    s += vis[b];
    ans.push(s);
  }
  return ans;
}