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Commit 561fc9f

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feat: add solutions to lc problem: No.2657 (#1967)
No.2657.Find the Prefix Common Array of Two Arrays
1 parent 6e704f2 commit 561fc9f

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7 files changed

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‎solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README.md‎

Lines changed: 108 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -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 。
4949

5050
<!-- 这里可写通用的实现逻辑 -->
5151

52-
**方法一:计数 + 枚举**
52+
**方法一:计数**
5353

5454
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。
5555

@@ -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 。
5959

6060
时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $A$ 和 $B$ 的长度。
6161

62+
**方法二:位运算(异或运算)**
63+
64+
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ドル$。另外,我们用一个变量 $s$ 记录当前公共元素的个数。
65+
66+
接下来,我们遍历数组 $A$ 和 $B,ドル更新 $vis[A[i]] = vis[A[i]] \oplus 1,ドル并且更新 $vis[B[i]] = vis[B[i]] \oplus 1,ドル其中 $\oplus$ 表示异或运算。
67+
68+
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ドル,ドル我们将 $s$ 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ドル,ドル我们将 $s$ 加一。然后将 $s$ 的值加入到答案数组 $ans$ 中。
69+
70+
遍历结束后,返回答案数组 $ans$ 即可。
71+
72+
时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $A$ 和 $B$ 的长度。
73+
6274
<!-- tabs:start -->
6375

6476
### **Python3**
@@ -79,6 +91,21 @@ class Solution:
7991
return ans
8092
```
8193

94+
```python
95+
class Solution:
96+
def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
97+
ans = []
98+
vis = [1] * (len(A) + 1)
99+
s = 0
100+
for a, b in zip(A, B):
101+
vis[a] ^= 1
102+
s += vis[a]
103+
vis[b] ^= 1
104+
s += vis[b]
105+
ans.append(s)
106+
return ans
107+
```
108+
82109
### **Java**
83110

84111
<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -102,6 +129,26 @@ class Solution {
102129
}
103130
```
104131

132+
```java
133+
class Solution {
134+
public int[] findThePrefixCommonArray(int[] A, int[] B) {
135+
int n = A.length;
136+
int[] ans = new int[n];
137+
int[] vis = new int[n + 1];
138+
Arrays.fill(vis, 1);
139+
int s = 0;
140+
for (int i = 0; i < n; ++i) {
141+
vis[A[i]] ^= 1;
142+
s += vis[A[i]];
143+
vis[B[i]] ^= 1;
144+
s += vis[B[i]];
145+
ans[i] = s;
146+
}
147+
return ans;
148+
}
149+
}
150+
```
151+
105152
### **C++**
106153

107154
```cpp
@@ -123,6 +170,26 @@ public:
123170
};
124171
```
125172
173+
```cpp
174+
class Solution {
175+
public:
176+
vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
177+
int n = A.size();
178+
vector<int> ans;
179+
vector<int> vis(n + 1, 1);
180+
int s = 0;
181+
for (int i = 0; i < n; ++i) {
182+
vis[A[i]] ^= 1;
183+
s += vis[A[i]];
184+
vis[B[i]] ^= 1;
185+
s += vis[B[i]];
186+
ans.push_back(s);
187+
}
188+
return ans;
189+
}
190+
};
191+
```
192+
126193
### **Go**
127194

128195
```go
@@ -143,14 +210,33 @@ func findThePrefixCommonArray(A []int, B []int) []int {
143210
}
144211
```
145212

213+
```go
214+
func findThePrefixCommonArray(A []int, B []int) (ans []int) {
215+
vis := make([]int, len(A)+1)
216+
for i := range vis {
217+
vis[i] = 1
218+
}
219+
s := 0
220+
for i, a := range A {
221+
b := B[i]
222+
vis[a] ^= 1
223+
s += vis[a]
224+
vis[b] ^= 1
225+
s += vis[b]
226+
ans = append(ans, s)
227+
}
228+
return
229+
}
230+
```
231+
146232
### **TypeScript**
147233

148234
```ts
149235
function findThePrefixCommonArray(A: number[], B: number[]): number[] {
150236
const n = A.length;
151-
const cnt1: number[] = newArray(n + 1).fill(0);
152-
const cnt2: number[] = newArray(n + 1).fill(0);
153-
const ans: number[] = newArray(n).fill(0);
237+
const cnt1: number[] = Array(n + 1).fill(0);
238+
const cnt2: number[] = Array(n + 1).fill(0);
239+
const ans: number[] = Array(n).fill(0);
154240
for (let i = 0; i < n; ++i) {
155241
++cnt1[A[i]];
156242
++cnt2[B[i]];
@@ -162,6 +248,24 @@ function findThePrefixCommonArray(A: number[], B: number[]): number[] {
162248
}
163249
```
164250

251+
```ts
252+
function findThePrefixCommonArray(A: number[], B: number[]): number[] {
253+
const n = A.length;
254+
const vis: number[] = Array(n + 1).fill(1);
255+
const ans: number[] = [];
256+
let s = 0;
257+
for (let i = 0; i < n; ++i) {
258+
const [a, b] = [A[i], B[i]];
259+
vis[a] ^= 1;
260+
s += vis[a];
261+
vis[b] ^= 1;
262+
s += vis[b];
263+
ans.push(s);
264+
}
265+
return ans;
266+
}
267+
```
268+
165269
### **...**
166270

167271
```

‎solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README_EN.md‎

Lines changed: 112 additions & 8 deletions
Original file line numberDiff line numberDiff line change
@@ -45,15 +45,27 @@ At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
4545

4646
## Solutions
4747

48-
**Solution 1: Count + Enumeration**
48+
**Solution 1: Counting**
4949

50-
We can use two arrays $cnt1$ and $cnt2$ to record the number of occurrences of each element in arrays $A$ and $B,ドル and use array $ans$ to record the answer.
50+
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.
5151

52-
Traverse arrays $A$ and $B,ドル 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],ドル calculate the minimum value of the number of occurrences of each element $j$ in $cnt1$ and $cnt2,ドル and add it to $ans[i]$.
52+
Traverse arrays $A$ and $B,ドル increment the occurrence times of $A[i]$ in $cnt1,ドル and increment the occurrence times of $B[i]$ in $cnt2$. Then enumerate $j \in [1,n],ドル calculate the minimum occurrence times of each element $j$ in $cnt1$ and $cnt2,ドル and accumulate them into $ans[i]$.
5353

54-
After the traversal is over, return the answer array $ans$.
54+
After the traversal, return the answer array $ans$.
5555

56-
The time complexity is $O(n^2),ドル and the space complexity is $O(n)$. Where $n$ is the length of arrays $A$ and $B$.
56+
The time complexity is $O(n^2),ドル and the space complexity is $O(n)$. Here, $n$ is the length of arrays $A$ and $B$.
57+
58+
**Solution 2: Bit Operation (XOR Operation)**
59+
60+
We can use an array $vis$ of length $n+1$ to record the occurrence situation of each element in arrays $A$ and $B,ドル the initial value of array $vis$ is 1ドル$. In addition, we use a variable $s$ to record the current number of common elements.
61+
62+
Next, we traverse arrays $A$ and $B,ドル update $vis[A[i]] = vis[A[i]] \oplus 1,ドル and update $vis[B[i]] = vis[B[i]] \oplus 1,ドル where $\oplus$ represents XOR operation.
63+
64+
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ドル,ドル and we increment $s$. Similarly, if the element $B[i]$ has appeared twice, then the value of $vis[B[i]]$ will be 1ドル,ドル and we increment $s$. Then add the value of $s$ to the answer array $ans$.
65+
66+
After the traversal, return the answer array $ans$.
67+
68+
The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of arrays $A$ and $B$.
5769

5870
<!-- tabs:start -->
5971

@@ -73,6 +85,21 @@ class Solution:
7385
return ans
7486
```
7587

88+
```python
89+
class Solution:
90+
def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
91+
ans = []
92+
vis = [1] * (len(A) + 1)
93+
s = 0
94+
for a, b in zip(A, B):
95+
vis[a] ^= 1
96+
s += vis[a]
97+
vis[b] ^= 1
98+
s += vis[b]
99+
ans.append(s)
100+
return ans
101+
```
102+
76103
### **Java**
77104

78105
```java
@@ -94,6 +121,26 @@ class Solution {
94121
}
95122
```
96123

124+
```java
125+
class Solution {
126+
public int[] findThePrefixCommonArray(int[] A, int[] B) {
127+
int n = A.length;
128+
int[] ans = new int[n];
129+
int[] vis = new int[n + 1];
130+
Arrays.fill(vis, 1);
131+
int s = 0;
132+
for (int i = 0; i < n; ++i) {
133+
vis[A[i]] ^= 1;
134+
s += vis[A[i]];
135+
vis[B[i]] ^= 1;
136+
s += vis[B[i]];
137+
ans[i] = s;
138+
}
139+
return ans;
140+
}
141+
}
142+
```
143+
97144
### **C++**
98145

99146
```cpp
@@ -115,6 +162,26 @@ public:
115162
};
116163
```
117164
165+
```cpp
166+
class Solution {
167+
public:
168+
vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
169+
int n = A.size();
170+
vector<int> ans;
171+
vector<int> vis(n + 1, 1);
172+
int s = 0;
173+
for (int i = 0; i < n; ++i) {
174+
vis[A[i]] ^= 1;
175+
s += vis[A[i]];
176+
vis[B[i]] ^= 1;
177+
s += vis[B[i]];
178+
ans.push_back(s);
179+
}
180+
return ans;
181+
}
182+
};
183+
```
184+
118185
### **Go**
119186

120187
```go
@@ -135,14 +202,33 @@ func findThePrefixCommonArray(A []int, B []int) []int {
135202
}
136203
```
137204

205+
```go
206+
func findThePrefixCommonArray(A []int, B []int) (ans []int) {
207+
vis := make([]int, len(A)+1)
208+
for i := range vis {
209+
vis[i] = 1
210+
}
211+
s := 0
212+
for i, a := range A {
213+
b := B[i]
214+
vis[a] ^= 1
215+
s += vis[a]
216+
vis[b] ^= 1
217+
s += vis[b]
218+
ans = append(ans, s)
219+
}
220+
return
221+
}
222+
```
223+
138224
### **TypeScript**
139225

140226
```ts
141227
function findThePrefixCommonArray(A: number[], B: number[]): number[] {
142228
const n = A.length;
143-
const cnt1: number[] = newArray(n + 1).fill(0);
144-
const cnt2: number[] = newArray(n + 1).fill(0);
145-
const ans: number[] = newArray(n).fill(0);
229+
const cnt1: number[] = Array(n + 1).fill(0);
230+
const cnt2: number[] = Array(n + 1).fill(0);
231+
const ans: number[] = Array(n).fill(0);
146232
for (let i = 0; i < n; ++i) {
147233
++cnt1[A[i]];
148234
++cnt2[B[i]];
@@ -154,6 +240,24 @@ function findThePrefixCommonArray(A: number[], B: number[]): number[] {
154240
}
155241
```
156242

243+
```ts
244+
function findThePrefixCommonArray(A: number[], B: number[]): number[] {
245+
const n = A.length;
246+
const vis: number[] = Array(n + 1).fill(1);
247+
const ans: number[] = [];
248+
let s = 0;
249+
for (let i = 0; i < n; ++i) {
250+
const [a, b] = [A[i], B[i]];
251+
vis[a] ^= 1;
252+
s += vis[a];
253+
vis[b] ^= 1;
254+
s += vis[b];
255+
ans.push(s);
256+
}
257+
return ans;
258+
}
259+
```
260+
157261
### **...**
158262

159263
```
Lines changed: 16 additions & 15 deletions
Original file line numberDiff line numberDiff line change
@@ -1,16 +1,17 @@
1-
class Solution {
2-
public:
3-
vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
4-
int n = A.size();
5-
vector<int> ans(n);
6-
vector<int> cnt1(n + 1), cnt2(n + 1);
7-
for (int i = 0; i < n; ++i) {
8-
++cnt1[A[i]];
9-
++cnt2[B[i]];
10-
for (int j = 1; j <= n; ++j) {
11-
ans[i] += min(cnt1[j], cnt2[j]);
12-
}
13-
}
14-
return ans;
15-
}
1+
class Solution {
2+
public:
3+
vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
4+
int n = A.size();
5+
vector<int> ans;
6+
vector<int> vis(n + 1, 1);
7+
int s = 0;
8+
for (int i = 0; i < n; ++i) {
9+
vis[A[i]] ^= 1;
10+
s += vis[A[i]];
11+
vis[B[i]] ^= 1;
12+
s += vis[B[i]];
13+
ans.push_back(s);
14+
}
15+
return ans;
16+
}
1617
};
Lines changed: 12 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,16 @@
1-
func findThePrefixCommonArray(A []int, B []int) []int {
2-
n := len(A)
3-
cnt1 := make([]int, n+1)
4-
cnt2 := make([]int, n+1)
5-
ans := make([]int, n)
1+
func findThePrefixCommonArray(A []int, B []int) (ans []int) {
2+
vis := make([]int, len(A)+1)
3+
for i := range vis {
4+
vis[i] = 1
5+
}
6+
s := 0
67
for i, a := range A {
78
b := B[i]
8-
cnt1[a]++
9-
cnt2[b]++
10-
forj:= 1; j<=n; j++ {
11-
ans[i] += min(cnt1[j], cnt2[j])
12-
}
9+
vis[a]^=1
10+
s+=vis[a]
11+
vis[b] ^= 1
12+
s += vis[b]
13+
ans=append(ans, s)
1314
}
14-
returnans
15+
return
1516
}

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