& nums) {
int ans = 0;
- for (int& v : nums) {
- ans += to_string(v).size() % 2 == 0;
+ for (int& x : nums) {
+ ans += to_string(x).size() % 2 == 0;
}
return ans;
}
@@ -112,8 +114,8 @@ public:
```go
func findNumbers(nums []int) (ans int) {
- for _, v := range nums {
- if len(strconv.Itoa(v))%2 == 0 {
+ for _, x := range nums {
+ if len(strconv.Itoa(x))%2 == 0 {
ans++
}
}
@@ -121,6 +123,14 @@ func findNumbers(nums []int) (ans int) {
}
```
+#### TypeScript
+
+```ts
+function findNumbers(nums: number[]): number {
+ return nums.filter(x => x.toString().length % 2 === 0).length;
+}
+```
+
#### JavaScript
```js
@@ -129,11 +139,7 @@ func findNumbers(nums []int) (ans int) {
* @return {number}
*/
var findNumbers = function (nums) {
- let ans = 0;
- for (const v of nums) {
- ans += String(v).length % 2 == 0;
- }
- return ans;
+ return nums.filter(x => x.toString().length % 2 === 0).length;
};
```
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md
index d9a7e8d0be679..2a92c8c6d1e4b 100644
--- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md
@@ -27,7 +27,7 @@ tags:
Input: nums = [12,345,2,6,7896]
Output: 2
-Explanation:
+Explanation:
12 contains 2 digits (even number of digits).
345 contains 3 digits (odd number of digits).
2 contains 1 digit (odd number of digits).
@@ -40,7 +40,7 @@ Therefore only 12 and 7896 contain an even number of digits.
Input: nums = [555,901,482,1771]
-Output: 1
+Output: 1
Explanation:
Only 1771 contains an even number of digits.
@@ -59,7 +59,13 @@ Only 1771 contains an even number of digits.
-### Solution 1
+### Solution 1: Simulation
+
+We traverse each element $x$ in the array $\textit{nums}$. For the current element $x$, we directly convert it to a string and then check if its length is even. If it is, we increment the answer by one.
+
+After the traversal is complete, we return the answer.
+
+The time complexity is $O(n \times \log M)$, and the space complexity is $O(\log M)$. Here, $n$ is the length of the array $\textit{nums}$, and $M$ is the maximum value of the elements in the array $\textit{nums}$.
@@ -68,7 +74,7 @@ Only 1771 contains an even number of digits.
```python
class Solution:
def findNumbers(self, nums: List[int]) -> int:
- return sum(len(str(v)) % 2 == 0 for v in nums)
+ return sum(len(str(x)) % 2 == 0 for x in nums)
```
#### Java
@@ -77,8 +83,8 @@ class Solution:
class Solution {
public int findNumbers(int[] nums) {
int ans = 0;
- for (int v : nums) {
- if (String.valueOf(v).length() % 2 == 0) {
+ for (int x : nums) {
+ if (String.valueOf(x).length() % 2 == 0) {
++ans;
}
}
@@ -94,8 +100,8 @@ class Solution {
public:
int findNumbers(vector& nums) {
int ans = 0;
- for (int& v : nums) {
- ans += to_string(v).size() % 2 == 0;
+ for (int& x : nums) {
+ ans += to_string(x).size() % 2 == 0;
}
return ans;
}
@@ -106,8 +112,8 @@ public:
```go
func findNumbers(nums []int) (ans int) {
- for _, v := range nums {
- if len(strconv.Itoa(v))%2 == 0 {
+ for _, x := range nums {
+ if len(strconv.Itoa(x))%2 == 0 {
ans++
}
}
@@ -115,6 +121,14 @@ func findNumbers(nums []int) (ans int) {
}
```
+#### TypeScript
+
+```ts
+function findNumbers(nums: number[]): number {
+ return nums.filter(x => x.toString().length % 2 === 0).length;
+}
+```
+
#### JavaScript
```js
@@ -123,11 +137,7 @@ func findNumbers(nums []int) (ans int) {
* @return {number}
*/
var findNumbers = function (nums) {
- let ans = 0;
- for (const v of nums) {
- ans += String(v).length % 2 == 0;
- }
- return ans;
+ return nums.filter(x => x.toString().length % 2 === 0).length;
};
```
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.cpp b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.cpp
index 01a83f17855b3..8a05d501250c3 100644
--- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.cpp
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.cpp
@@ -2,9 +2,9 @@ class Solution {
public:
int findNumbers(vector& nums) {
int ans = 0;
- for (int& v : nums) {
- ans += to_string(v).size() % 2 == 0;
+ for (int& x : nums) {
+ ans += to_string(x).size() % 2 == 0;
}
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.go b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.go
index df389d4b63993..2b714feabcb8c 100644
--- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.go
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.go
@@ -1,8 +1,8 @@
func findNumbers(nums []int) (ans int) {
- for _, v := range nums {
- if len(strconv.Itoa(v))%2 == 0 {
+ for _, x := range nums {
+ if len(strconv.Itoa(x))%2 == 0 {
ans++
}
}
return
-}
\ No newline at end of file
+}
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.java b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.java
index 5bd0d8d0ef99a..41e9941e31cf6 100644
--- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.java
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.java
@@ -1,11 +1,11 @@
class Solution {
public int findNumbers(int[] nums) {
int ans = 0;
- for (int v : nums) {
- if (String.valueOf(v).length() % 2 == 0) {
+ for (int x : nums) {
+ if (String.valueOf(x).length() % 2 == 0) {
++ans;
}
}
return ans;
}
-}
\ No newline at end of file
+}
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.js b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.js
index ca37b182c22ab..5b77e8f4a2efa 100644
--- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.js
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.js
@@ -3,9 +3,5 @@
* @return {number}
*/
var findNumbers = function (nums) {
- let ans = 0;
- for (const v of nums) {
- ans += String(v).length % 2 == 0;
- }
- return ans;
+ return nums.filter(x => x.toString().length % 2 === 0).length;
};
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.py b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.py
index 6d311f5596c2b..5fd671760e1eb 100644
--- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.py
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.py
@@ -1,3 +1,3 @@
class Solution:
def findNumbers(self, nums: List[int]) -> int:
- return sum(len(str(v)) % 2 == 0 for v in nums)
+ return sum(len(str(x)) % 2 == 0 for x in nums)
diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.ts b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.ts
new file mode 100644
index 0000000000000..dee6e231b5acc
--- /dev/null
+++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.ts
@@ -0,0 +1,3 @@
+function findNumbers(nums: number[]): number {
+ return nums.filter(x => x.toString().length % 2 === 0).length;
+}
diff --git a/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md b/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md
index 900f63d6ad0c9..c2b1874c84663 100644
--- a/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md
+++ b/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md
@@ -75,9 +75,9 @@ tags:
### 方法一：哈希表 + 枚举
-根据题目描述，如果一个长串满足条件，那么这个长串的子串（长度至少为 `minSize`）也一定满足条件。因此，我们只需要枚举 $s$ 中所有长度为 `minSize` 的子串，然后利用哈希表记录所有子串的出现次数，找出最大的次数作为答案即可。
+根据题目描述，如果一个长串满足条件，那么这个长串的长度为 $\textit{minSize}$ 的子串也一定满足条件。因此，我们只需要枚举 $s$ 中所有长度为 $\textit{minSize}$ 的子串，然后利用哈希表记录所有子串的出现次数，找出最大的次数作为答案即可。
-时间复杂度 $O(n \times m)$，空间复杂度 $O(n \times m)$。其中 $n$ 和 $m$ 分别为字符串 $s$ 的长度以及 `minSize` 的大小。本题中 $m$ 不超过 $26$。
+时间复杂度 $O(n \times m)$，空间复杂度 $O(n \times m)$。其中 $n$ 和 $m$ 分别为字符串 $s$ 的长度以及 $\textit{minSize}$ 的大小。本题中 $m$ 不超过 $26$。
@@ -162,6 +162,24 @@ func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
}
```
+#### TypeScript
+
+```ts
+function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
+ const cnt = new Map();
+ let ans = 0;
+ for (let i = 0; i < s.length - minSize + 1; ++i) {
+ const t = s.slice(i, i + minSize);
+ const ss = new Set(t.split(''));
+ if (ss.size <= maxLetters) {
+ cnt.set(t, (cnt.get(t) || 0) + 1);
+ ans = Math.max(ans, cnt.get(t)!);
+ }
+ }
+ return ans;
+}
+```
+
diff --git a/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md b/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md
index 9a8a00bcdce57..2aa30af79cc2e 100644
--- a/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md
+++ b/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md
@@ -61,7 +61,11 @@ It satisfies the conditions, 2 unique letters and size 3 (between minSize and ma
-### Solution 1
+### Solution 1: Hash Table + Enumeration
+
+According to the problem description, if a long string meets the condition, then its substring of length $\textit{minSize}$ must also meet the condition. Therefore, we only need to enumerate all substrings of length $\textit{minSize}$ in $s$, then use a hash table to record the occurrence frequency of all substrings, and find the maximum frequency as the answer.
+
+The time complexity is $O(n \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the lengths of the string $s$ and $\textit{minSize}$, respectively. In this problem, $m$ does not exceed $26$.
@@ -146,6 +150,24 @@ func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
}
```
+#### TypeScript
+
+```ts
+function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
+ const cnt = new Map();
+ let ans = 0;
+ for (let i = 0; i < s.length - minSize + 1; ++i) {
+ const t = s.slice(i, i + minSize);
+ const ss = new Set(t.split(''));
+ if (ss.size <= maxLetters) {
+ cnt.set(t, (cnt.get(t) || 0) + 1);
+ ans = Math.max(ans, cnt.get(t)!);
+ }
+ }
+ return ans;
+}
+```
+
diff --git a/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.ts b/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.ts
new file mode 100644
index 0000000000000..2654afd8e4b3e
--- /dev/null
+++ b/solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.ts
@@ -0,0 +1,13 @@
+function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
+ const cnt = new Map();
+ let ans = 0;
+ for (let i = 0; i < s.length - minSize + 1; ++i) {
+ const t = s.slice(i, i + minSize);
+ const ss = new Set(t.split(''));
+ if (ss.size <= maxLetters) {
+ cnt.set(t, (cnt.get(t) || 0) + 1);
+ ans = Math.max(ans, cnt.get(t)!);
+ }
+ }
+ return ans;
+}
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README.md b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README.md
index 19c111f488a6b..feb5314f203a3 100644
--- a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README.md
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README.md
@@ -61,7 +61,15 @@ tags:
-### 方法一
+### 方法一：逆序遍历
+
+我们用一个变量 $mx$ 记录当前位置右侧的最大值，初始时 $mx = -1$。
+
+然后我们从右向左遍历数组，对于每个位置 $i$，我们记当前位置的值为 $x$，将当前位置的值更新为 $mx$，然后更新 $mx = \max(mx, x)$。
+
+最后返回原数组即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
@@ -70,11 +78,11 @@ tags:
```python
class Solution:
def replaceElements(self, arr: List[int]) -> List[int]:
- m = -1
- for i in range(len(arr) - 1, -1, -1):
- t = arr[i]
- arr[i] = m
- m = max(m, t)
+ mx = -1
+ for i in reversed(range(len(arr))):
+ x = arr[i]
+ arr[i] = mx
+ mx = max(mx, x)
return arr
```
@@ -83,13 +91,55 @@ class Solution:
```java
class Solution {
public int[] replaceElements(int[] arr) {
- for (int i = arr.length - 1, max = -1; i >= 0; --i) {
- int t = arr[i];
- arr[i] = max;
- max = Math.max(max, t);
+ for (int i = arr.length - 1, mx = -1; i >= 0; --i) {
+ int x = arr[i];
+ arr[i] = mx;
+ mx = Math.max(mx, x);
+ }
+ return arr;
+ }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+ vector replaceElements(vector& arr) {
+ for (int i = arr.size() - 1, mx = -1; ~i; --i) {
+ int x = arr[i];
+ arr[i] = mx;
+ mx = max(mx, x);
}
return arr;
}
+};
+```
+
+#### Go
+
+```go
+func replaceElements(arr []int) []int {
+ for i, mx := len(arr)-1, -1; i >= 0; i-- {
+ x := arr[i]
+ arr[i] = mx
+ mx = max(mx, x)
+ }
+ return arr
+}
+```
+
+#### TypeScript
+
+```ts
+function replaceElements(arr: number[]): number[] {
+ for (let i = arr.length - 1, mx = -1; ~i; --i) {
+ const x = arr[i];
+ arr[i] = mx;
+ mx = Math.max(mx, x);
+ }
+ return arr;
}
```
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md
index c6101fce18df1..5cd11d3f3f03c 100644
--- a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md
@@ -28,7 +28,7 @@ tags:
Input: arr = [17,18,5,4,6,1]
Output: [18,6,6,6,1,-1]
-Explanation:
+Explanation:
- index 0 --> the greatest element to the right of index 0 is index 1 (18).
- index 1 --> the greatest element to the right of index 1 is index 4 (6).
- index 2 --> the greatest element to the right of index 2 is index 4 (6).
@@ -59,7 +59,15 @@ tags:
-### Solution 1
+### Solution 1: Reverse Traversal
+
+We use a variable $mx$ to record the maximum value to the right of the current position, initially $mx = -1$.
+
+Then we traverse the array from right to left. For each position $i$, we denote the current value as $x$, update the current position's value to $mx$, and then update $mx = \max(mx, x)$.
+
+Finally, return the original array.
+
+The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
@@ -68,11 +76,11 @@ tags:
```python
class Solution:
def replaceElements(self, arr: List[int]) -> List[int]:
- m = -1
- for i in range(len(arr) - 1, -1, -1):
- t = arr[i]
- arr[i] = m
- m = max(m, t)
+ mx = -1
+ for i in reversed(range(len(arr))):
+ x = arr[i]
+ arr[i] = mx
+ mx = max(mx, x)
return arr
```
@@ -81,13 +89,55 @@ class Solution:
```java
class Solution {
public int[] replaceElements(int[] arr) {
- for (int i = arr.length - 1, max = -1; i >= 0; --i) {
- int t = arr[i];
- arr[i] = max;
- max = Math.max(max, t);
+ for (int i = arr.length - 1, mx = -1; i >= 0; --i) {
+ int x = arr[i];
+ arr[i] = mx;
+ mx = Math.max(mx, x);
+ }
+ return arr;
+ }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+ vector replaceElements(vector& arr) {
+ for (int i = arr.size() - 1, mx = -1; ~i; --i) {
+ int x = arr[i];
+ arr[i] = mx;
+ mx = max(mx, x);
}
return arr;
}
+};
+```
+
+#### Go
+
+```go
+func replaceElements(arr []int) []int {
+ for i, mx := len(arr)-1, -1; i >= 0; i-- {
+ x := arr[i]
+ arr[i] = mx
+ mx = max(mx, x)
+ }
+ return arr
+}
+```
+
+#### TypeScript
+
+```ts
+function replaceElements(arr: number[]): number[] {
+ for (let i = arr.length - 1, mx = -1; ~i; --i) {
+ const x = arr[i];
+ arr[i] = mx;
+ mx = Math.max(mx, x);
+ }
+ return arr;
}
```
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.cpp b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.cpp
new file mode 100644
index 0000000000000..faac001515a5d
--- /dev/null
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.cpp
@@ -0,0 +1,11 @@
+class Solution {
+public:
+ vector replaceElements(vector& arr) {
+ for (int i = arr.size() - 1, mx = -1; ~i; --i) {
+ int x = arr[i];
+ arr[i] = mx;
+ mx = max(mx, x);
+ }
+ return arr;
+ }
+};
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.go b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.go
new file mode 100644
index 0000000000000..fa1282ac18a52
--- /dev/null
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.go
@@ -0,0 +1,8 @@
+func replaceElements(arr []int) []int {
+ for i, mx := len(arr)-1, -1; i >= 0; i-- {
+ x := arr[i]
+ arr[i] = mx
+ mx = max(mx, x)
+ }
+ return arr
+}
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.java b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.java
index 8d7b95db66a4b..5260011ea070a 100644
--- a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.java
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.java
@@ -1,10 +1,10 @@
class Solution {
public int[] replaceElements(int[] arr) {
- for (int i = arr.length - 1, max = -1; i >= 0; --i) {
- int t = arr[i];
- arr[i] = max;
- max = Math.max(max, t);
+ for (int i = arr.length - 1, mx = -1; i >= 0; --i) {
+ int x = arr[i];
+ arr[i] = mx;
+ mx = Math.max(mx, x);
}
return arr;
}
-}
\ No newline at end of file
+}
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.py b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.py
index bc3fa6cc61bd4..b67306f30ab59 100644
--- a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.py
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.py
@@ -1,8 +1,8 @@
class Solution:
def replaceElements(self, arr: List[int]) -> List[int]:
- m = -1
- for i in range(len(arr) - 1, -1, -1):
- t = arr[i]
- arr[i] = m
- m = max(m, t)
+ mx = -1
+ for i in reversed(range(len(arr))):
+ x = arr[i]
+ arr[i] = mx
+ mx = max(mx, x)
return arr
diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.ts b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.ts
new file mode 100644
index 0000000000000..c5773e8ac6748
--- /dev/null
+++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/Solution.ts
@@ -0,0 +1,8 @@
+function replaceElements(arr: number[]): number[] {
+ for (let i = arr.length - 1, mx = -1; ~i; --i) {
+ const x = arr[i];
+ arr[i] = mx;
+ mx = Math.max(mx, x);
+ }
+ return arr;
+}
diff --git a/solution/3100-3199/3162.Find the Number of Good Pairs I/README.md b/solution/3100-3199/3162.Find the Number of Good Pairs I/README.md
index bf962a213512e..3f749bf5b7988 100644
--- a/solution/3100-3199/3162.Find the Number of Good Pairs I/README.md
+++ b/solution/3100-3199/3162.Find the Number of Good Pairs I/README.md
@@ -69,11 +69,11 @@ tags:
### 方法一：暴力枚举
-我们直接枚举所有的数位 $(x, y)$，判断是否满足 $x \mod (y \times k) = 0$，如果满足则答案加一。
+我们直接枚举所有的数位 $(x, y)$，判断是否满足 $x \bmod (y \times k) = 0$，如果满足则答案加一。
枚举结束后，返回答案即可。
-时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是数组 `nums1` 和 `nums2` 的长度。空间复杂度 $O(1)$。
+时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是数组 $\textit{nums1}$ 和 $\textit{nums2}$ 的长度。空间复杂度 $O(1)$。
diff --git a/solution/3100-3199/3162.Find the Number of Good Pairs I/README_EN.md b/solution/3100-3199/3162.Find the Number of Good Pairs I/README_EN.md
index c7da5c2e19a70..fc90667fd61c3 100644
--- a/solution/3100-3199/3162.Find the Number of Good Pairs I/README_EN.md
+++ b/solution/3100-3199/3162.Find the Number of Good Pairs I/README_EN.md
@@ -65,11 +65,11 @@ The 5 good pairs are (0, 0), (1, 0), (1, 1)