feat: add solutions to lc problem: No.1477 (#4409)

No.1477.Find Two Non-overlapping Sub-arrays Each With Target Sum

Author: yanglbme

solution/1400-1499/1477.Find Two Non-overlapping Sub-arrays Each With Target Sum/README.md

@@ -85,9 +85,9 @@ tags:
 
 我们可以使用哈希表 $d$ 记录前缀和最近一次出现的位置，初始时 $d[0]=0$。
 
-定义 $f[i]$ 表示前 $i$ 个元素中，长度和为 $target$ 的最短子数组的长度。初始时 $f[0]=inf$。
+定义 $f[i]$ 表示前 $i$ 个元素中，长度和为 $target$ 的最短子数组的长度。初始时 $f[0]= \infty$。
 
-遍历数组 `arr`，对于当前位置 $i$，计算前缀和 $s$，如果 $s-target$ 在哈希表中，记 $j=d[s-target]$，则 $f[i]=min(f[i],i-j)$，答案为 $ans=min(ans,f[j]+i-j)$。继续遍历下个位置。
+遍历数组 $\textit{arr}$，对于当前位置 $i$，计算前缀和 $s$，如果 $s - \textit{target}$ 在哈希表中，记 $j=d[s - \textit{target}]$，则 $f[i]=\min(f[i], i - j)$，答案为 $ans=\min(ans, f[j] + i - j)$。继续遍历下个位置。
 
 最后，如果答案大于数组长度，则返回 $-1$，否则返回答案。
 
@@ -199,6 +199,33 @@ func minSumOfLengths(arr []int, target int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function minSumOfLengths(arr: number[], target: number): number {
+    const d = new Map<number, number>();
+    d.set(0, 0);
+    let s = 0;
+    const n = arr.length;
+    const f: number[] = Array(n + 1);
+    const inf = 1 << 30;
+    f[0] = inf;
+    let ans = inf;
+    for (let i = 1; i <= n; ++i) {
+        const v = arr[i - 1];
+        s += v;
+        f[i] = f[i - 1];
+        if (d.has(s - target)) {
+            const j = d.get(s - target)!;
+            f[i] = Math.min(f[i], i - j);
+            ans = Math.min(ans, f[j] + i - j);
+        }
+        d.set(s, i);
+    }
+    return ans > n ? -1 : ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1400-1499/1477.Find Two Non-overlapping Sub-arrays Each With Target Sum/README_EN.md

@@ -68,7 +68,17 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Hash Table + Prefix Sum + Dynamic Programming
+
+We can use a hash table $d$ to record the most recent position where each prefix sum appears, with the initial value $d[0]=0$.
+
+Define $f[i]$ as the minimum length of a subarray with sum equal to $target$ among the first $i$ elements. Initially, $f[0]=\infty$.
+
+Iterate through the array $\textit{arr}$. For the current position $i$, calculate the prefix sum $s$. If $s - \textit{target}$ exists in the hash table, let $j = d[s - \textit{target}]$, then $f[i] = \min(f[i], i - j)$, and the answer is $ans = \min(ans, f[j] + i - j)$. Continue to the next position.
+
+Finally, if the answer is greater than the array length, return $-1$; otherwise, return the answer.
+
+The complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array.
 
 <!-- tabs:start -->
 
@@ -176,6 +186,33 @@ func minSumOfLengths(arr []int, target int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function minSumOfLengths(arr: number[], target: number): number {
+    const d = new Map<number, number>();
+    d.set(0, 0);
+    let s = 0;
+    const n = arr.length;
+    const f: number[] = Array(n + 1);
+    const inf = 1 << 30;
+    f[0] = inf;
+    let ans = inf;
+    for (let i = 1; i <= n; ++i) {
+        const v = arr[i - 1];
+        s += v;
+        f[i] = f[i - 1];
+        if (d.has(s - target)) {
+            const j = d.get(s - target)!;
+            f[i] = Math.min(f[i], i - j);
+            ans = Math.min(ans, f[j] + i - j);
+        }
+        d.set(s, i);
+    }
+    return ans > n ? -1 : ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1400-1499/1477.Find Two Non-overlapping Sub-arrays Each With Target Sum/Solution.ts

@@ -0,0 +1,22 @@
+function minSumOfLengths(arr: number[], target: number): number {
+    const d = new Map<number, number>();
+    d.set(0, 0);
+    let s = 0;
+    const n = arr.length;
+    const f: number[] = Array(n + 1);
+    const inf = 1 << 30;
+    f[0] = inf;
+    let ans = inf;
+    for (let i = 1; i <= n; ++i) {
+        const v = arr[i - 1];
+        s += v;
+        f[i] = f[i - 1];
+        if (d.has(s - target)) {
+            const j = d.get(s - target)!;
+            f[i] = Math.min(f[i], i - j);
+            ans = Math.min(ans, f[j] + i - j);
+        }
+        d.set(s, i);
+    }
+    return ans > n ? -1 : ans;
+}

solution/1400-1499/1487.Making File Names Unique/README_EN.md

@@ -63,9 +63,9 @@ tags:
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= names.length &lt;= 5 * 10<sup>4</sup></code></li>
-	<li><code>1 &lt;= names[i].length &lt;= 20</code></li>
-	<li><code>names[i]</code> consists of lowercase English letters, digits, and/or round brackets.</li>
+ <li><code>1 &lt;= names.length &lt;= 5 * 10<sup>4</sup></code></li>
+ <li><code>1 &lt;= names[i].length &lt;= 20</code></li>
+ <li><code>names[i]</code> consists of lowercase English letters, digits, and/or round brackets.</li>
 </ul>
 
 <!-- description:end -->
@@ -74,7 +74,21 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Hash Table
+
+We can use a hash table $d$ to record the minimum available index for each folder name, where $d[name] = k$ means the minimum available index for the folder $name$ is $k$. Initially, $d$ is empty since there are no folders.
+
+Next, we iterate through the folder names array. For each file name $name$:
+
+-   If $name$ is already in $d$, it means the folder $name$ already exists, and we need to find a new folder name. We can keep trying $name(k)$, where $k$ starts from $d[name]$, until we find a folder name $name(k)$ that does not exist in $d$. We add $name(k)$ to $d$, update $d[name]$ to $k + 1$, and then update $name$ to $name(k)$.
+-   If $name$ is not in $d$, we can directly add $name$ to $d$ and set $d[name]$ to $1$.
+-   Then, we add $name$ to the answer array and continue to the next file name.
+
+After traversing all file names, we obtain the answer array.
+
+> In the code implementation below, we directly modify the $names$ array without using an extra answer array.
+
+The complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the sum of the lengths of all file names in the $names$ array.
 
 <!-- tabs:start -->
 
@@ -144,22 +158,22 @@ public:
 
 ```go
 func getFolderNames(names []string) []string {
-	d := map[string]int{}
-	for i, name := range names {
-		if k, ok := d[name]; ok {
-			for {
-				newName := fmt.Sprintf("%s(%d)", name, k)
-				if d[newName] == 0 {
-					d[name] = k + 1
-					names[i] = newName
-					break
-				}
-				k++
-			}
-		}
-		d[names[i]] = 1
-	}
-	return names
+ d := map[string]int{}
+ for i, name := range names {
+  if k, ok := d[name]; ok {
+   for {
+    newName := fmt.Sprintf("%s(%d)", name, k)
+    if d[newName] == 0 {
+     d[name] = k + 1
+     names[i] = newName
+     break
+    }
+    k++
+   }
+  }
+  d[names[i]] = 1
+ }
+ return names
 }
 ```