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Commit b335300

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feat: add solutions to lc problems: No.1295,1299 (doocs#3629)
* No.1295.Find Numbers with Even Number of Digits * No.1297.Maximum Number of Occurrences of a Substring * No.1299.Replace Elements with Greatest Element on Right Side
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‎solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md‎

Lines changed: 22 additions & 16 deletions
Original file line numberDiff line numberDiff line change
@@ -41,7 +41,7 @@ tags:
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<pre>
4343
<strong>输入:</strong>nums = [555,901,482,1771]
44-
<strong>输出:</strong>1
44+
<strong>输出:</strong>1
4545
<strong>解释: </strong>
4646
只有 1771 是位数为偶数的数字。
4747
</pre>
@@ -61,11 +61,13 @@ tags:
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<!-- solution:start -->
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64-
### 方法一:枚举
64+
### 方法一:模拟
6565

66-
枚举数组 `nums` 中的每个元素,将其转换为字符串,判断字符串长度是否为偶数,是则答案加一。
66+
我们遍历数组 $\textit{nums}$ 中的每个元素,对于当前遍历到的元素 $x,ドル我们直接将其转换为字符串,然后判断其长度是否为偶数即可。若是则将答案加一。
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68-
时间复杂度 $O(n \times \log_{10} m),ドル空间复杂度 $O(\log_{10} m),ドル其中 $n$ 和 $m$ 分别为数组 `nums` 的长度以及数组 `nums` 中的最大元素。
68+
遍历结束后,我们返回答案即可。
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时间复杂度 $O(n \times \log M),ドル空间复杂度 $O(\log M)$。其中 $n$ 是数组 $\textit{nums}$ 的长度,而 $M$ 是数组 $\textit{nums}$ 中的元素的最大值。
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<!-- tabs:start -->
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@@ -74,7 +76,7 @@ tags:
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```python
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class Solution:
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def findNumbers(self, nums: List[int]) -> int:
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return sum(len(str(v)) % 2 == 0 for v in nums)
79+
return sum(len(str(x)) % 2 == 0 for x in nums)
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```
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#### Java
@@ -83,8 +85,8 @@ class Solution:
8385
class Solution {
8486
public int findNumbers(int[] nums) {
8587
int ans = 0;
86-
for (int v : nums) {
87-
if (String.valueOf(v).length() % 2 == 0) {
88+
for (int x : nums) {
89+
if (String.valueOf(x).length() % 2 == 0) {
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++ans;
8991
}
9092
}
@@ -100,8 +102,8 @@ class Solution {
100102
public:
101103
int findNumbers(vector<int>& nums) {
102104
int ans = 0;
103-
for (int& v : nums) {
104-
ans += to_string(v).size() % 2 == 0;
105+
for (int& x : nums) {
106+
ans += to_string(x).size() % 2 == 0;
105107
}
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return ans;
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}
@@ -112,15 +114,23 @@ public:
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113115
```go
114116
func findNumbers(nums []int) (ans int) {
115-
for _, v := range nums {
116-
if len(strconv.Itoa(v))%2 == 0 {
117+
for _, x := range nums {
118+
if len(strconv.Itoa(x))%2 == 0 {
117119
ans++
118120
}
119121
}
120122
return
121123
}
122124
```
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#### TypeScript
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```ts
129+
function findNumbers(nums: number[]): number {
130+
return nums.filter(x => x.toString().length % 2 === 0).length;
131+
}
132+
```
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#### JavaScript
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```js
@@ -129,11 +139,7 @@ func findNumbers(nums []int) (ans int) {
129139
* @return {number}
130140
*/
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var findNumbers = function (nums) {
132-
let ans = 0;
133-
for (const v of nums) {
134-
ans += String(v).length % 2 == 0;
135-
}
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return ans;
142+
return nums.filter(x => x.toString().length % 2 === 0).length;
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};
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```
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‎solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md‎

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<pre>
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<strong>Input:</strong> nums = [12,345,2,6,7896]
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<strong>Output:</strong> 2
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<strong>Explanation:
30+
<strong>Explanation:
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</strong>12 contains 2 digits (even number of digits).&nbsp;
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345 contains 3 digits (odd number of digits).&nbsp;
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2 contains 1 digit (odd number of digits).&nbsp;
@@ -40,7 +40,7 @@ Therefore only 12 and 7896 contain an even number of digits.
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<pre>
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<strong>Input:</strong> nums = [555,901,482,1771]
43-
<strong>Output:</strong> 1
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<strong>Output:</strong> 1
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<strong>Explanation: </strong>
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Only 1771 contains an even number of digits.
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</pre>
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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We traverse each element $x$ in the array $\textit{nums}$. For the current element $x,ドル we directly convert it to a string and then check if its length is even. If it is, we increment the answer by one.
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After the traversal is complete, we return the answer.
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The time complexity is $O(n \times \log M),ドル and the space complexity is $O(\log M)$. Here, $n$ is the length of the array $\textit{nums},ドル and $M$ is the maximum value of the elements in the array $\textit{nums}$.
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<!-- tabs:start -->
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```python
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class Solution:
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def findNumbers(self, nums: List[int]) -> int:
71-
return sum(len(str(v)) % 2 == 0 for v in nums)
77+
return sum(len(str(x)) % 2 == 0 for x in nums)
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```
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#### Java
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7783
class Solution {
7884
public int findNumbers(int[] nums) {
7985
int ans = 0;
80-
for (int v : nums) {
81-
if (String.valueOf(v).length() % 2 == 0) {
86+
for (int x : nums) {
87+
if (String.valueOf(x).length() % 2 == 0) {
8288
++ans;
8389
}
8490
}
@@ -94,8 +100,8 @@ class Solution {
94100
public:
95101
int findNumbers(vector<int>& nums) {
96102
int ans = 0;
97-
for (int& v : nums) {
98-
ans += to_string(v).size() % 2 == 0;
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for (int& x : nums) {
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ans += to_string(x).size() % 2 == 0;
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}
100106
return ans;
101107
}
@@ -106,15 +112,23 @@ public:
106112
107113
```go
108114
func findNumbers(nums []int) (ans int) {
109-
for _, v := range nums {
110-
if len(strconv.Itoa(v))%2 == 0 {
115+
for _, x := range nums {
116+
if len(strconv.Itoa(x))%2 == 0 {
111117
ans++
112118
}
113119
}
114120
return
115121
}
116122
```
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124+
#### TypeScript
125+
126+
```ts
127+
function findNumbers(nums: number[]): number {
128+
return nums.filter(x => x.toString().length % 2 === 0).length;
129+
}
130+
```
131+
118132
#### JavaScript
119133

120134
```js
@@ -123,11 +137,7 @@ func findNumbers(nums []int) (ans int) {
123137
* @return {number}
124138
*/
125139
var findNumbers = function (nums) {
126-
let ans = 0;
127-
for (const v of nums) {
128-
ans += String(v).length % 2 == 0;
129-
}
130-
return ans;
140+
return nums.filter(x => x.toString().length % 2 === 0).length;
131141
};
132142
```
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‎solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.cpp‎

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Original file line numberDiff line numberDiff line change
@@ -2,9 +2,9 @@ class Solution {
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public:
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int findNumbers(vector<int>& nums) {
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int ans = 0;
5-
for (int& v : nums) {
6-
ans += to_string(v).size() % 2 == 0;
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for (int& x : nums) {
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ans += to_string(x).size() % 2 == 0;
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}
88
return ans;
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}
10-
};
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};
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func findNumbers(nums []int) (ans int) {
2-
for _, v := range nums {
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if len(strconv.Itoa(v))%2 == 0 {
2+
for _, x := range nums {
3+
if len(strconv.Itoa(x))%2 == 0 {
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ans++
55
}
66
}
77
return
8-
}
8+
}
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Original file line numberDiff line numberDiff line change
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11
class Solution {
22
public int findNumbers(int[] nums) {
33
int ans = 0;
4-
for (int v : nums) {
5-
if (String.valueOf(v).length() % 2 == 0) {
4+
for (int x : nums) {
5+
if (String.valueOf(x).length() % 2 == 0) {
66
++ans;
77
}
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}
99
return ans;
1010
}
11-
}
11+
}

‎solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.js‎

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Original file line numberDiff line numberDiff line change
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33
* @return {number}
44
*/
55
var findNumbers = function (nums) {
6-
let ans = 0;
7-
for (const v of nums) {
8-
ans += String(v).length % 2 == 0;
9-
}
10-
return ans;
6+
return nums.filter(x => x.toString().length % 2 === 0).length;
117
};
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Original file line numberDiff line numberDiff line change
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11
class Solution:
22
def findNumbers(self, nums: List[int]) -> int:
3-
return sum(len(str(v)) % 2 == 0 for v in nums)
3+
return sum(len(str(x)) % 2 == 0 for x in nums)
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,3 @@
1+
function findNumbers(nums: number[]): number {
2+
return nums.filter(x => x.toString().length % 2 === 0).length;
3+
}

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md‎

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### 方法一:哈希表 + 枚举
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78-
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。
78+
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。
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80-
时间复杂度 $O(n \times m),ドル空间复杂度 $O(n \times m)$。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ドル$。
80+
时间复杂度 $O(n \times m),ドル空间复杂度 $O(n \times m)$。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ドル$。
8181

8282
<!-- tabs:start -->
8383

@@ -162,6 +162,24 @@ func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
162162
}
163163
```
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165+
#### TypeScript
166+
167+
```ts
168+
function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
169+
const cnt = new Map<string, number>();
170+
let ans = 0;
171+
for (let i = 0; i < s.length - minSize + 1; ++i) {
172+
const t = s.slice(i, i + minSize);
173+
const ss = new Set(t.split(''));
174+
if (ss.size <= maxLetters) {
175+
cnt.set(t, (cnt.get(t) || 0) + 1);
176+
ans = Math.max(ans, cnt.get(t)!);
177+
}
178+
}
179+
return ans;
180+
}
181+
```
182+
165183
<!-- tabs:end -->
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167185
<!-- solution:end -->

‎solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -61,7 +61,11 @@ It satisfies the conditions, 2 unique letters and size 3 (between minSize and ma
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<!-- solution:start -->
6363

64-
### Solution 1
64+
### Solution 1: Hash Table + Enumeration
65+
66+
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,ドル then use a hash table to record the occurrence frequency of all substrings, and find the maximum frequency as the answer.
67+
68+
The time complexity is $O(n \times m),ドル and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the lengths of the string $s$ and $\textit{minSize},ドル respectively. In this problem, $m$ does not exceed 26ドル$.
6569

6670
<!-- tabs:start -->
6771

@@ -146,6 +150,24 @@ func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
146150
}
147151
```
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153+
#### TypeScript
154+
155+
```ts
156+
function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
157+
const cnt = new Map<string, number>();
158+
let ans = 0;
159+
for (let i = 0; i < s.length - minSize + 1; ++i) {
160+
const t = s.slice(i, i + minSize);
161+
const ss = new Set(t.split(''));
162+
if (ss.size <= maxLetters) {
163+
cnt.set(t, (cnt.get(t) || 0) + 1);
164+
ans = Math.max(ans, cnt.get(t)!);
165+
}
166+
}
167+
return ans;
168+
}
169+
```
170+
149171
<!-- tabs:end -->
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151173
<!-- solution:end -->

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