feat: compelete No.1043

Author: BaffinLee

1001-1100/1043. Partition Array for Maximum Sum.md

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+# 1043. Partition Array for Maximum Sum
+
+- Difficulty: Medium.
+- Related Topics: Array, Dynamic Programming.
+- Similar Questions: Subsequence of Size K With the Largest Even Sum.
+
+## Problem
+
+Given an integer array ```arr```, partition the array into (contiguous) subarrays of length **at most** ```k```. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
+
+Return **the largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a **32-bit** integer.**
+
+ 
+Example 1:
+
+```
+Input: arr = [1,15,7,9,2,5,10], k = 3
+Output: 84
+Explanation: arr becomes [15,15,15,9,10,10,10]
+```
+
+Example 2:
+
+```
+Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
+Output: 83
+```
+
+Example 3:
+
+```
+Input: arr = [1], k = 1
+Output: 1
+```
+
+ 
+**Constraints:**
+
+
+	
+- ```1 <= arr.length <= 500```
+	
+- ```0 <= arr[i] <= 109```
+	
+- ```1 <= k <= arr.length```
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} arr
+ * @param {number} k
+ * @return {number}
+ */
+var maxSumAfterPartitioning = function(arr, k) {
+    var dp = Array(arr.length).fill(0);
+    for (var i = 0; i < arr.length; i++) {
+        var maxValue = 0;
+        for (var j = 1; j <= k && j - 1 <= i; j++) {
+            maxValue = Math.max(maxValue, arr[i - j + 1]);
+            dp[i] = Math.max(dp[i], ( dp[i - j + 1] + maxValue * j);
+        }
+    }
+    return dp[arr.length];
+};
+```
+
+**Explain:**
+
+`dp[i + 1]` represents array `[0 ... i]` 's maxSumAfterPartitioning. (dp[0] = 0 means nothing, it just helps calculate.)
+
+```
+dp[i] = max(
+  dp[i - 1] + max(arr[i]) * 1,
+  dp[i - 2] + max(arr[i], arr[i - 1]) * 2,
+  ...
+  dp[i - k] + max(arr[i], arr[i - 1], ..., arr[i - k]) * k
+)
+```
+
+`dp[arr.length - 1]` would be the answer in the end of the iteration.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).