feat: add solutions to lc problems: No.1749,1750

* No.1749.Maximum Absolute Sum of Any Subarray
* No.1750.Minimum Length of String After Deleting Similar Ends

Author: yanglbme

solution/1700-1799/1749.Maximum Absolute Sum of Any Subarray/README.md

@@ -155,6 +155,22 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function maxAbsoluteSum(nums: number[]): number {
+    let f = 0;
+    let g = 0;
+    let ans = 0;
+    for (const x of nums) {
+        f = Math.max(f, 0) + x;
+        g = Math.min(g, 0) + x;
+        ans = Math.max(ans, f, -g);
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```

solution/1700-1799/1749.Maximum Absolute Sum of Any Subarray/README_EN.md

@@ -42,6 +42,23 @@
 
 ## Solutions
 
+**Approach 1: Dynamic Programming**
+
+We define $f[i]$ to represent the maximum value of the subarray ending with $nums[i]$, and define $g[i]$ to represent the minimum value of the subarray ending with $nums[i]$. Then the state transition equation of $f[i]$ and $g[i]$ is as follows:
+
+$$
+\begin{aligned}
+f[i] &= \max(f[i - 1], 0) + nums[i] \\
+g[i] &= \min(g[i - 1], 0) + nums[i]
+\end{aligned}
+$$
+
+The final answer is the maximum value of $max(f[i], |g[i]|)$.
+
+Since $f[i]$ and $g[i]$ are only related to $f[i - 1]$ and $g[i - 1]$, we can use two variables to replace the array, reducing the space complexity to $O(1)$.
+
+Time complexity $O(n)$, space complexity $O(1)$, where $n$ is the length of the array $nums$.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -128,6 +145,22 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function maxAbsoluteSum(nums: number[]): number {
+    let f = 0;
+    let g = 0;
+    let ans = 0;
+    for (const x of nums) {
+        f = Math.max(f, 0) + x;
+        g = Math.min(g, 0) + x;
+        ans = Math.max(ans, f, -g);
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```

solution/1700-1799/1749.Maximum Absolute Sum of Any Subarray/Solution.ts

@@ -0,0 +1,11 @@
+function maxAbsoluteSum(nums: number[]): number {
+    let f = 0;
+    let g = 0;
+    let ans = 0;
+    for (const x of nums) {
+        f = Math.max(f, 0) + x;
+        g = Math.min(g, 0) + x;
+        ans = Math.max(ans, f, -g);
+    }
+    return ans;
+}

solution/1700-1799/1750.Minimum Length of String After Deleting Similar Ends/README.md

@@ -161,20 +161,19 @@ func max(a, b int) int {
 
 ```ts
 function minimumLength(s: string): number {
-    const n = s.length;
-    let start = 0;
-    let end = n - 1;
-    while (start < end && s[start] === s[end]) {
-        while (start + 1 < end && s[start] === s[start + 1]) {
-            start++;
+    let i = 0;
+    let j = s.length - 1;
+    while (i < j && s[i] === s[j]) {
+        while (i + 1 < j && s[i + 1] === s[i]) {
+            ++i;
         }
-        while (start < end - 1 && s[end] === s[end - 1]) {
-            end--;
+        while (i < j - 1 && s[j - 1] === s[j]) {
+            --j;
         }
-        start++;
-        end--;
+        ++i;
+        --j;
     }
-    return Math.max(0, end - start + 1);
+    return Math.max(0, j - i + 1);
 }
 ```
 

solution/1700-1799/1750.Minimum Length of String After Deleting Similar Ends/README_EN.md

@@ -56,6 +56,12 @@
 
 ## Solutions
 
+**Approach 1: Two pointers**
+
+We define two pointers $i$ and $j$ to point to the head and tail of the string $s$ respectively, then move them to the middle until the characters pointed to by $i$ and $j$ are not equal, then $\max(0, j - i + 1)$ is the answer.
+
+The time complexity is $O(n)$ and the space complexity is $O(1)$. Where $n$ is the length of the string $s$.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -145,20 +151,19 @@ func max(a, b int) int {
 
 ```ts
 function minimumLength(s: string): number {
-    const n = s.length;
-    let start = 0;
-    let end = n - 1;
-    while (start < end && s[start] === s[end]) {
-        while (start + 1 < end && s[start] === s[start + 1]) {
-            start++;
+    let i = 0;
+    let j = s.length - 1;
+    while (i < j && s[i] === s[j]) {
+        while (i + 1 < j && s[i + 1] === s[i]) {
+            ++i;
         }
-        while (start < end - 1 && s[end] === s[end - 1]) {
-            end--;
+        while (i < j - 1 && s[j - 1] === s[j]) {
+            --j;
         }
-        start++;
-        end--;
+        ++i;
+        --j;
     }
-    return Math.max(0, end - start + 1);
+    return Math.max(0, j - i + 1);
 }
 ```
 

solution/1700-1799/1750.Minimum Length of String After Deleting Similar Ends/Solution.ts

@@ -1,16 +1,15 @@
 function minimumLength(s: string): number {
-    const n = s.length;
-    let start = 0;
-    let end = n - 1;
-    while (start < end && s[start] === s[end]) {
-        while (start + 1 < end && s[start] === s[start + 1]) {
-            start++;
+    let i = 0;
+    let j = s.length - 1;
+    while (i < j && s[i] === s[j]) {
+        while (i + 1 < j && s[i + 1] === s[i]) {
+            ++i;
         }
-        while (start < end - 1 && s[end] === s[end - 1]) {
-            end--;
+        while (i < j - 1 && s[j - 1] === s[j]) {
+            --j;
         }
-        start++;
-        end--;
+        ++i;
+        --j;
     }
-    return Math.max(0, end - start + 1);
+    return Math.max(0, j - i + 1);
 }