feat: add solutions to lc problems: No.0643,1984 (#2556)

* No.0643.Maximum Average Subarray I
* No.1984.Minimum Difference Between Highest and Lowest of K Scores

Author: yanglbme

solution/0600-0699/0643.Maximum Average Subarray I/README.md

@@ -43,15 +43,18 @@
 
 ## 解法
 
-### 方法一
+### 方法一：滑动窗口
+
+我们维护一个长度为 $k$ 的滑动窗口，每次计算窗口内的和 $s$，取最大的和 $s$ 作为答案。
+
+时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def findMaxAverage(self, nums: List[int], k: int) -> float:
-        s = sum(nums[:k])
-        ans = s
+        ans = s = sum(nums[:k])
         for i in range(k, len(nums)):
             s += nums[i] - nums[i - k]
             ans = max(ans, s)
@@ -75,19 +78,46 @@ class Solution {
 }
 ```
 
+```cpp
+class Solution {
+public:
+    double findMaxAverage(vector<int>& nums, int k) {
+        int s = accumulate(nums.begin(), nums.begin() + k, 0);
+        int ans = s;
+        for (int i = k; i < nums.size(); ++i) {
+            s += nums[i] - nums[i - k];
+            ans = max(ans, s);
+        }
+        return static_cast<double>(ans) / k;
+    }
+};
+```
+
+```go
+func findMaxAverage(nums []int, k int) float64 {
+	s := 0
+	for _, x := range nums[:k] {
+		s += x
+	}
+	ans := s
+	for i := k; i < len(nums); i++ {
+		s += nums[i] - nums[i-k]
+		ans = max(ans, s)
+	}
+	return float64(ans) / float64(k)
+}
+```
+
 ```ts
 function findMaxAverage(nums: number[], k: number): number {
-    let n = nums.length;
-    let ans = 0;
-    let sum = 0;
-    // 前k
-    for (let i = 0; i < k; i++) {
-        sum += nums[i];
+    let s = 0;
+    for (let i = 0; i < k; ++i) {
+        s += nums[i];
     }
-    ans = sum;
-    for (let i = k; i < n; i++) {
-        sum += nums[i] - nums[i - k];
-        ans = Math.max(ans, sum);
+    let ans = s;
+    for (let i = k; i < nums.length; ++i) {
+        s += nums[i] - nums[i - k];
+        ans = Math.max(ans, s);
     }
     return ans / k;
 }
@@ -98,13 +128,13 @@ impl Solution {
     pub fn find_max_average(nums: Vec<i32>, k: i32) -> f64 {
         let k = k as usize;
         let n = nums.len();
-        let mut sum = nums.iter().take(k).sum::<i32>();
-        let mut max = sum;
+        let mut s = nums.iter().take(k).sum::<i32>();
+        let mut ans = s;
         for i in k..n {
-            sum += nums[i] - nums[i - k];
-            max = max.max(sum);
+            s += nums[i] - nums[i - k];
+            ans = ans.max(s);
         }
-        f64::from(max) / f64::from(k as i32)
+        f64::from(ans) / f64::from(k as i32)
     }
 }
 ```
@@ -117,16 +147,16 @@ class Solution {
      * @return Float
      */
     function findMaxAverage($nums, $k) {
-        $sum = 0;
+        $s = 0;
         for ($i = 0; $i < $k; $i++) {
-            $sum += $nums[$i];
+            $s += $nums[$i];
         }
-        $max = $sum;
+        $ans = $s;
         for ($j = $k; $j < count($nums); $j++) {
-            $sum = $sum - $nums[$j - $k] + $nums[$j];
-            $max = max($max, $sum);
+            $s = $s - $nums[$j - $k] + $nums[$j];
+            $ans = max($ans, $s);
         }
-        return $max / $k;
+        return $ans / $k;
     }
 }
 ```

solution/0600-0699/0643.Maximum Average Subarray I/README_EN.md

@@ -37,15 +37,18 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Sliding Window
+
+We maintain a sliding window of length $k$, and for each window, we calculate the sum $s$ of the numbers within the window. We take the maximum sum $s$ as the answer.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def findMaxAverage(self, nums: List[int], k: int) -> float:
-        s = sum(nums[:k])
-        ans = s
+        ans = s = sum(nums[:k])
         for i in range(k, len(nums)):
             s += nums[i] - nums[i - k]
             ans = max(ans, s)
@@ -69,19 +72,46 @@ class Solution {
 }
 ```
 
+```cpp
+class Solution {
+public:
+    double findMaxAverage(vector<int>& nums, int k) {
+        int s = accumulate(nums.begin(), nums.begin() + k, 0);
+        int ans = s;
+        for (int i = k; i < nums.size(); ++i) {
+            s += nums[i] - nums[i - k];
+            ans = max(ans, s);
+        }
+        return static_cast<double>(ans) / k;
+    }
+};
+```
+
+```go
+func findMaxAverage(nums []int, k int) float64 {
+	s := 0
+	for _, x := range nums[:k] {
+		s += x
+	}
+	ans := s
+	for i := k; i < len(nums); i++ {
+		s += nums[i] - nums[i-k]
+		ans = max(ans, s)
+	}
+	return float64(ans) / float64(k)
+}
+```
+
 ```ts
 function findMaxAverage(nums: number[], k: number): number {
-    let n = nums.length;
-    let ans = 0;
-    let sum = 0;
-    // 前k
-    for (let i = 0; i < k; i++) {
-        sum += nums[i];
+    let s = 0;
+    for (let i = 0; i < k; ++i) {
+        s += nums[i];
     }
-    ans = sum;
-    for (let i = k; i < n; i++) {
-        sum += nums[i] - nums[i - k];
-        ans = Math.max(ans, sum);
+    let ans = s;
+    for (let i = k; i < nums.length; ++i) {
+        s += nums[i] - nums[i - k];
+        ans = Math.max(ans, s);
     }
     return ans / k;
 }
@@ -92,13 +122,13 @@ impl Solution {
     pub fn find_max_average(nums: Vec<i32>, k: i32) -> f64 {
         let k = k as usize;
         let n = nums.len();
-        let mut sum = nums.iter().take(k).sum::<i32>();
-        let mut max = sum;
+        let mut s = nums.iter().take(k).sum::<i32>();
+        let mut ans = s;
         for i in k..n {
-            sum += nums[i] - nums[i - k];
-            max = max.max(sum);
+            s += nums[i] - nums[i - k];
+            ans = ans.max(s);
         }
-        f64::from(max) / f64::from(k as i32)
+        f64::from(ans) / f64::from(k as i32)
     }
 }
 ```
@@ -111,16 +141,16 @@ class Solution {
      * @return Float
      */
     function findMaxAverage($nums, $k) {
-        $sum = 0;
+        $s = 0;
         for ($i = 0; $i < $k; $i++) {
-            $sum += $nums[$i];
+            $s += $nums[$i];
         }
-        $max = $sum;
+        $ans = $s;
         for ($j = $k; $j < count($nums); $j++) {
-            $sum = $sum - $nums[$j - $k] + $nums[$j];
-            $max = max($max, $sum);
+            $s = $s - $nums[$j - $k] + $nums[$j];
+            $ans = max($ans, $s);
         }
-        return $max / $k;
+        return $ans / $k;
     }
 }
 ```

solution/0600-0699/0643.Maximum Average Subarray I/Solution.cpp

@@ -0,0 +1,12 @@
+class Solution {
+public:
+    double findMaxAverage(vector<int>& nums, int k) {
+        int s = accumulate(nums.begin(), nums.begin() + k, 0);
+        int ans = s;
+        for (int i = k; i < nums.size(); ++i) {
+            s += nums[i] - nums[i - k];
+            ans = max(ans, s);
+        }
+        return static_cast<double>(ans) / k;
+    }
+};

solution/0600-0699/0643.Maximum Average Subarray I/Solution.go

@@ -0,0 +1,12 @@
+func findMaxAverage(nums []int, k int) float64 {
+	s := 0
+	for _, x := range nums[:k] {
+		s += x
+	}
+	ans := s
+	for i := k; i < len(nums); i++ {
+		s += nums[i] - nums[i-k]
+		ans = max(ans, s)
+	}
+	return float64(ans) / float64(k)
+}

solution/0600-0699/0643.Maximum Average Subarray I/Solution.php

@@ -5,15 +5,15 @@ class Solution {
      * @return Float
      */
     function findMaxAverage($nums, $k) {
-        $sum = 0;
+        $s = 0;
         for ($i = 0; $i < $k; $i++) {
-            $sum += $nums[$i];
+            $s += $nums[$i];
         }
-        $max = $sum;
+        $ans = $s;
         for ($j = $k; $j < count($nums); $j++) {
-            $sum = $sum - $nums[$j - $k] + $nums[$j];
-            $max = max($max, $sum);
+            $s = $s - $nums[$j - $k] + $nums[$j];
+            $ans = max($ans, $s);
         }
-        return $max / $k;
+        return $ans / $k;
     }
 }

solution/0600-0699/0643.Maximum Average Subarray I/Solution.py

@@ -1,7 +1,6 @@
 class Solution:
     def findMaxAverage(self, nums: List[int], k: int) -> float:
-        s = sum(nums[:k])
-        ans = s
+        ans = s = sum(nums[:k])
         for i in range(k, len(nums)):
             s += nums[i] - nums[i - k]
             ans = max(ans, s)

solution/0600-0699/0643.Maximum Average Subarray I/Solution.rs

@@ -2,12 +2,12 @@ impl Solution {
     pub fn find_max_average(nums: Vec<i32>, k: i32) -> f64 {
         let k = k as usize;
         let n = nums.len();
-        let mut sum = nums.iter().take(k).sum::<i32>();
-        let mut max = sum;
+        let mut s = nums.iter().take(k).sum::<i32>();
+        let mut ans = s;
         for i in k..n {
-            sum += nums[i] - nums[i - k];
-            max = max.max(sum);
+            s += nums[i] - nums[i - k];
+            ans = ans.max(s);
         }
-        f64::from(max) / f64::from(k as i32)
+        f64::from(ans) / f64::from(k as i32)
     }
 }

solution/0600-0699/0643.Maximum Average Subarray I/Solution.ts

@@ -1,15 +1,12 @@
 function findMaxAverage(nums: number[], k: number): number {
-    let n = nums.length;
-    let ans = 0;
-    let sum = 0;
-    // 前k
-    for (let i = 0; i < k; i++) {
-        sum += nums[i];
+    let s = 0;
+    for (let i = 0; i < k; ++i) {
+        s += nums[i];
     }
-    ans = sum;
-    for (let i = k; i < n; i++) {
-        sum += nums[i] - nums[i - k];
-        ans = Math.max(ans, sum);
+    let ans = s;
+    for (let i = k; i < nums.length; ++i) {
+        s += nums[i] - nums[i - k];
+        ans = Math.max(ans, s);
     }
     return ans / k;
 }

solution/0600-0699/0644.Maximum Average Subarray II/README_EN.md

@@ -42,7 +42,41 @@ Note that we do not consider the subarrays of length < 4.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Binary Search
+
+We note that if the average value of a subarray with length greater than or equal to $k$ is $v$, then the maximum average number must be greater than or equal to $v$, otherwise the maximum average number must be less than $v$. Therefore, we can use binary search to find the maximum average number.
+
+What are the left and right boundaries of binary search? The left boundary $l$ must be the minimum value in the array, and the right boundary $r$ is the maximum value in the array. Next, we binary search the midpoint $mid$, and judge whether there exists a subarray with length greater than or equal to $k$ whose average value is greater than or equal to $mid$. If it exists, then we update the left boundary $l$ to $mid$, otherwise we update the right boundary $r$ to $mid$. When the difference between the left and right boundaries is less than a very small non-negative number, i.e., $r - l < \epsilon$, we can get the maximum average number, where $\epsilon$ represents a very small positive number, which can be $10^{-5}$.
+
+The key to the problem is how to judge whether the average value of a subarray with length greater than or equal to $k$ is greater than or equal to $v$.
+
+We assume that in the array $nums$, there is a subarray with length $j$, the elements are $a_1, a_2, \cdots, a_j$, and its average value is greater than or equal to $v$, i.e.,
+
+$$
+\frac{a_1 + a_2 + \cdots + a_j}{j} \geq v
+$$
+
+Then,
+
+$$
+a_1 + a_2 + \cdots + a_j \geq v \times j
+$$
+
+That is,
+
+$$
+(a_1 - v) + (a_2 - v) + \cdots + (a_j - v) \geq 0
+$$
+
+We can find that if we subtract $v$ from each element in the array $nums$, the original problem is transformed into a problem of whether the sum of the elements of a subarray with length greater than or equal to $k$ is greater than or equal to $0$. We can use a sliding window to solve this problem.
+
+First, we calculate the sum $s$ of the differences between the first $k$ elements and $v$. If $s \geq 0$, it means that there exists a subarray with length greater than or equal to $k$ whose element sum is greater than or equal to $0$.
+
+Otherwise, we continue to traverse the element $nums[j]$. Suppose the current sum of the differences between the first $j$ elements and $v$ is $s_j$. Then we can maintain the minimum value $mi$ of the sum of the differences between the prefix sum and $v$ in the range $[0,..j-k]$. If $s_j \geq mi$ exists, it means that there exists a subarray with length greater than or equal to $k$ whose element sum is greater than or equal to $0$, and we return $true$.
+
+Otherwise, we continue to traverse the element $nums[j]$ until the entire array is traversed.
+
+The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length of the array $nums$ and the difference between the maximum and minimum values in the array, respectively. The space complexity is $O(1)$.
 
 <!-- tabs:start -->