feat: update solutions to lc problems: No.1508,1512 (doocs#3579)

Author: yanglbme

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README.md

@@ -31,7 +31,7 @@ tags:
 
 <pre>
 <strong>输入：</strong>nums = [1,2,3,4], n = 4, left = 1, right = 5
-<strong>输出：</strong>13 
+<strong>输出：</strong>13
 <strong>解释：</strong>所有的子数组和为 1, 3, 6, 10, 2, 5, 9, 3, 7, 4 。将它们升序排序后，我们得到新的数组 [1, 2, 3, 3, 4, 5, 6, 7, 9, 10] 。下标从 le = 1 到 ri = 5 的和为 1 + 2 + 3 + 3 + 4 = 13 。
 </pre>
 
@@ -67,11 +67,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一：排序
+### 方法一：模拟
 
-按照题意生成 `arr` 数组，排序后，对 $[left-1,.. right-1]$ 范围的所有元素求和，得到结果。
+我们可以按照题目的要求，生成数组 $\textit{arr}$，然后对数组进行排序，最后求出 $[\textit{left}-1, \textit{right}-1]$ 范围的所有元素的和，得到结果。
 
-时间复杂度 $O(n^2\log n)$，空间复杂度 $O(n^2)$。其中 $n$ 为题目给定的数组长度。
+时间复杂度 $O(n^2 \times \log n)$，空间复杂度 $O(n^2)$。其中 $n$ 为题目给定的数组长度。
 
 <!-- tabs:start -->
 
@@ -175,13 +175,8 @@ function rangeSum(nums: number[], n: number, left: number, right: number): numbe
         }
     }
 
-    let ans = 0;
     arr = arr.sort((a, b) => a - b).slice(left - 1, right);
-    for (const x of arr) {
-        ans += x;
-    }
-
-    return ans % mod;
+    return arr.reduce((acc, cur) => (acc + cur) % mod, 0);
 }
 ```
 
@@ -199,13 +194,8 @@ function rangeSum(nums, n, left, right) {
         }
     }
 
-    let ans = 0;
     arr = arr.sort((a, b) => a - b).slice(left - 1, right);
-    for (const x of arr) {
-        ans += x;
-    }
-
-    return ans % mod;
+    return arr.reduce((acc, cur) => acc + cur, 0) % mod;
 }
 ```
 

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README_EN.md

@@ -30,8 +30,8 @@ tags:
 
 <pre>
 <strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 1, right = 5
-<strong>Output:</strong> 13 
-<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 
+<strong>Output:</strong> 13
+<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
@@ -65,11 +65,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1: Sorting
+### Solution 1: Simulation
 
-According to the problem statement, generate the `arr` array, sort it, and then sum all the elements in the range $[left-1,.. right-1]$ to get the result.
+We can generate the array $\textit{arr}$ according to the problem's requirements, then sort the array, and finally calculate the sum of all elements in the range $[\textit{left}-1, \textit{right}-1]$ to get the result.
 
-Time complexity is $O(n^2 \times \log n)$, and space complexity is $O(n^2)$. Here, $n$ is the length of the array given in the problem.
+The time complexity is $O(n^2 \times \log n)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array given in the problem.
 
 <!-- tabs:start -->
 
@@ -173,13 +173,8 @@ function rangeSum(nums: number[], n: number, left: number, right: number): numbe
         }
     }
 
-    let ans = 0;
     arr = arr.sort((a, b) => a - b).slice(left - 1, right);
-    for (const x of arr) {
-        ans += x;
-    }
-
-    return ans % mod;
+    return arr.reduce((acc, cur) => (acc + cur) % mod, 0);
 }
 ```
 
@@ -197,13 +192,8 @@ function rangeSum(nums, n, left, right) {
         }
     }
 
-    let ans = 0;
     arr = arr.sort((a, b) => a - b).slice(left - 1, right);
-    for (const x of arr) {
-        ans += x;
-    }
-
-    return ans % mod;
+    return arr.reduce((acc, cur) => acc + cur, 0) % mod;
 }
 ```
 

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/Solution.js

@@ -9,11 +9,6 @@ function rangeSum(nums, n, left, right) {
         }
     }
 
-    let ans = 0;
     arr = arr.sort((a, b) => a - b).slice(left - 1, right);
-    for (const x of arr) {
-        ans += x;
-    }
-
-    return ans % mod;
+    return arr.reduce((acc, cur) => acc + cur, 0) % mod;
 }

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/Solution.ts

@@ -9,11 +9,6 @@ function rangeSum(nums: number[], n: number, left: number, right: number): numbe
         }
     }
 
-    let ans = 0;
     arr = arr.sort((a, b) => a - b).slice(left - 1, right);
-    for (const x of arr) {
-        ans += x;
-    }
-
-    return ans % mod;
+    return arr.reduce((acc, cur) => (acc + cur) % mod, 0);
 }

solution/1500-1599/1512.Number of Good Pairs/README.md

@@ -132,7 +132,7 @@ func numIdenticalPairs(nums []int) (ans int) {
 
 ```ts
 function numIdenticalPairs(nums: number[]): number {
-    const cnt = new Array(101).fill(0);
+    const cnt: number[] = Array(101).fill(0);
     let ans = 0;
     for (const x of nums) {
         ans += cnt[x]++;
@@ -146,17 +146,34 @@ function numIdenticalPairs(nums: number[]): number {
 ```rust
 impl Solution {
     pub fn num_identical_pairs(nums: Vec<i32>) -> i32 {
-        let mut cnt = [0; 101];
         let mut ans = 0;
-        for &num in nums.iter() {
-            ans += cnt[num as usize];
-            cnt[num as usize] += 1;
+        let mut cnt = [0; 101];
+        for &x in nums.iter() {
+            ans += cnt[x as usize];
+            cnt[x as usize] += 1;
         }
         ans
     }
 }
 ```
 
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var numIdenticalPairs = function (nums) {
+    const cnt = Array(101).fill(0);
+    let ans = 0;
+    for (const x of nums) {
+        ans += cnt[x]++;
+    }
+    return ans;
+};
+```
+
 #### PHP
 
 ```php
@@ -166,15 +183,12 @@ class Solution {
      * @return Integer
      */
     function numIdenticalPairs($nums) {
-        $arr = array_values(array_unique($nums));
-        for ($i = 0; $i < count($nums); $i++) {
-            $v[$nums[$i]] += 1;
+        $ans = 0;
+        $cnt = array_fill(0, 101, 0);
+        foreach ($nums as $x) {
+            $ans += $cnt[$x]++;
         }
-        $rs = 0;
-        for ($j = 0; $j < count($arr); $j++) {
-            $rs += ($v[$arr[$j]] * ($v[$arr[$j]] - 1)) / 2;
-        }
-        return $rs;
+        return $ans;
     }
 }
 ```
@@ -196,125 +210,4 @@ int numIdenticalPairs(int* nums, int numsSize) {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def numIdenticalPairs(self, nums: List[int]) -> int:
-        cnt = Counter(nums)
-        return sum(v * (v - 1) for v in cnt.values()) >> 1
-```
-
-#### Java
-
-```java
-class Solution {
-    public int numIdenticalPairs(int[] nums) {
-        int[] cnt = new int[101];
-        for (int x : nums) {
-            ++cnt[x];
-        }
-        int ans = 0;
-        for (int v : cnt) {
-            ans += v * (v - 1) / 2;
-        }
-        return ans;
-    }
-}
-```
-
-#### C++
-
-```cpp
-class Solution {
-public:
-    int numIdenticalPairs(vector<int>& nums) {
-        int cnt[101]{};
-        for (int& x : nums) {
-            ++cnt[x];
-        }
-        int ans = 0;
-        for (int v : cnt) {
-            ans += v * (v - 1) / 2;
-        }
-        return ans;
-    }
-};
-```
-
-#### Go
-
-```go
-func numIdenticalPairs(nums []int) (ans int) {
-	cnt := [101]int{}
-	for _, x := range nums {
-		cnt[x]++
-	}
-	for _, v := range cnt {
-		ans += v * (v - 1) / 2
-	}
-	return
-}
-```
-
-#### TypeScript
-
-```ts
-function numIdenticalPairs(nums: number[]): number {
-    const cnt = new Array(101).fill(0);
-    for (const x of nums) {
-        ++cnt[x];
-    }
-    let ans = 0;
-    for (const v of cnt) {
-        ans += v * (v - 1);
-    }
-    return ans >> 1;
-}
-```
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn num_identical_pairs(nums: Vec<i32>) -> i32 {
-        let mut cnt = [0; 101];
-        for &num in nums.iter() {
-            cnt[num as usize] += 1;
-        }
-        let mut ans = 0;
-        for &v in cnt.iter() {
-            ans += (v * (v - 1)) / 2;
-        }
-        ans
-    }
-}
-```
-
-#### C
-
-```c
-int numIdenticalPairs(int* nums, int numsSize) {
-    int cnt[101] = {0};
-    for (int i = 0; i < numsSize; i++) {
-        cnt[nums[i]]++;
-    }
-    int ans = 0;
-    for (int i = 0; i < 101; ++i) {
-        ans += cnt[i] * (cnt[i] - 1) / 2;
-    }
-    return ans;
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->