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Nov 15, 2023
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feat: add solutions to lc problem: No.2657 #1967

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112 changes: 108 additions & 4 deletions solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -49,7 +49,7 @@ i = 2：1，2 和 3 是两个数组的前缀公共元素，所以 C[2] = 3 。

<!-- 这里可写通用的实现逻辑 -->

**方法一：计数 + 枚举**
**方法一：计数**

我们可以使用两个数组 $cnt1$ 和 $cnt2$ 分别记录数组 $A$ 和 $B$ 中每个元素出现的次数，用数组 $ans$ 记录答案。

Expand All @@ -59,6 +59,18 @@ i = 2：1，2 和 3 是两个数组的前缀公共元素，所以 C[2] = 3 。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $A$ 和 $B$ 的长度。

**方法二：位运算（异或运算）**

我们可以使用一个长度为 $n+1$ 的数组 $vis$ 记录数组 $A$ 和 $B$ 中每个元素的出现情况，数组 $vis$ 的初始值为 $1$。另外，我们用一个变量 $s$ 记录当前公共元素的个数。

接下来，我们遍历数组 $A$ 和 $B$，更新 $vis[A[i]] = vis[A[i]] \oplus 1$，并且更新 $vis[B[i]] = vis[B[i]] \oplus 1$，其中 $\oplus$ 表示异或运算。

如果遍历到当前位置，元素 $A[i]$ 出现过两次（即在数组 $A$ 和 $B$ 中都出现过），那么 $vis[A[i]]$ 的值将为会 $1$，我们将 $s$ 加一。同理，如果元素 $B[i]$ 出现过两次，那么 $vis[B[i]]$ 的值将为会 $1$，我们将 $s$ 加一。然后将 $s$ 的值加入到答案数组 $ans$ 中。

遍历结束后，返回答案数组 $ans$ 即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $A$ 和 $B$ 的长度。

<!-- tabs:start -->

### **Python3**
Expand All @@ -79,6 +91,21 @@ class Solution:
return ans
```

```python
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
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->
Expand All @@ -102,6 +129,26 @@ class Solution {
}
```

```java
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;
}
}
```

### **C++**

```cpp
Expand All @@ -123,6 +170,26 @@ public:
};
```

```cpp
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;
}
};
```

### **Go**

```go
Expand All @@ -143,14 +210,33 @@ func findThePrefixCommonArray(A []int, B []int) []int {
}
```

```go
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
}
```

### **TypeScript**

```ts
function findThePrefixCommonArray(A: number[], B: number[]): number[] {
const n = A.length;
const cnt1: number[] = new Array(n + 1).fill(0);
const cnt2: number[] = new Array(n + 1).fill(0);
const ans: number[] = new Array(n).fill(0);
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]];
Expand All @@ -162,6 +248,24 @@ function findThePrefixCommonArray(A: number[], B: number[]): number[] {
}
```

```ts
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;
}
```

### **...**

```
Expand Down
120 changes: 112 additions & 8 deletions solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -45,15 +45,27 @@ At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.

## Solutions

**Solution 1: Count + Enumeration**
**Solution 1: Counting**

We can use two arrays $cnt1$ and $cnt2$ to record the number of occurrences of each element in arrays $A$ and $B$, and use array $ans$ to record the answer.
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 $B$, add the number of occurrences of $A[i]$ in $cnt1$ by one, and add the number of occurrences of $B[i]$ in $cnt2$ by one. Then enumerate $j \in [1,n]$, calculate the minimum value of the number of occurrences of each element $j$ in $cnt1$ and $cnt2$, and add it to $ans[i]$.
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 is over, return the answer array $ans$.
After the traversal, return the answer array $ans$.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the length of arrays $A$ and $B$.
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 $B$.

<!-- tabs:start -->

Expand All @@ -73,6 +85,21 @@ class Solution:
return ans
```

```python
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
```

### **Java**

```java
Expand All @@ -94,6 +121,26 @@ class Solution {
}
```

```java
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;
}
}
```

### **C++**

```cpp
Expand All @@ -115,6 +162,26 @@ public:
};
```

```cpp
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;
}
};
```

### **Go**

```go
Expand All @@ -135,14 +202,33 @@ func findThePrefixCommonArray(A []int, B []int) []int {
}
```

```go
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
}
```

### **TypeScript**

```ts
function findThePrefixCommonArray(A: number[], B: number[]): number[] {
const n = A.length;
const cnt1: number[] = new Array(n + 1).fill(0);
const cnt2: number[] = new Array(n + 1).fill(0);
const ans: number[] = new Array(n).fill(0);
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]];
Expand All @@ -154,6 +240,24 @@ function findThePrefixCommonArray(A: number[], B: number[]): number[] {
}
```

```ts
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;
}
```

### **...**

```
Expand Down
31 changes: 16 additions & 15 deletions solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,16 +1,17 @@
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 {
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;
}
};
23 changes: 12 additions & 11 deletions solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,15 +1,16 @@
func findThePrefixCommonArray(A []int, B []int) []int {
n := len(A)
cnt1 := make([]int, n+1)
cnt2 := make([]int, n+1)
ans := make([]int, n)
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]
cnt1[a]++
cnt2[b]++
for j := 1; j <= n; j++ {
ans[i] += min(cnt1[j], cnt2[j])
}
vis[a] ^= 1
s += vis[a]
vis[b] ^= 1
s += vis[b]
ans = append(ans, s)
}
return ans
return
}
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