feat: add solutions to lc problem: No.1590

solution/1500-1599/1590.Make Sum Divisible by P/README.md

@@ -80,17 +80,17 @@ tags:
 
 ### 方法一：前缀和 + 哈希表
 
-我们可以先求出数组 $nums$ 所有元素之和模 $p$ 的值，记为 $k$。如果 $k$ 为 $0$，说明数组 $nums$ 所有元素之和就是 $p$ 的倍数，直接返回 $0$ 即可。
+我们可以先求出数组 $\textit{nums}$ 所有元素之和模 $p$ 的值，记为 $k$。如果 $k$ 为 $0$，说明数组 $\textit{nums}$ 所有元素之和就是 $p$ 的倍数，直接返回 $0$ 即可。
 
 如果 $k$ 不为 $0$，我们需要找到一个最短的子数组，使得删除该子数组后，剩余元素之和模 $p$ 的值为 $0$。
 
-我们可以遍历数组 $nums$，维护当前的前缀和模 $p$ 的值，记为 $cur$。用哈希表 $last$ 记录每个前缀和模 $p$ 的值最后一次出现的位置。
+我们可以遍历数组 $\textit{nums}$，维护当前的前缀和模 $p$ 的值，记为 $cur$。用哈希表 $last$ 记录每个前缀和模 $p$ 的值最后一次出现的位置。
 
-如果当前存在一个以 $nums[i]$ 结尾的子数组，使得删除该子数组后，剩余元素之和模 $p$ 的值为 $0$。也就是说，我们需要找到此前的一个前缀和模 $p$ 的值为 $target$ 的位置 $j$，使得 $(target + k - cur) \bmod p = 0$。如果找到，我们就可以将 $j + 1$ 到 $i$ 这一段闭区间子数组 $nums[j+1,..i]$ 删除，使得剩余元素之和模 $p$ 的值为 $0$。
+如果当前存在一个以 $\textit{nums}[i]$ 结尾的子数组，使得删除该子数组后，剩余元素之和模 $p$ 的值为 $0$。也就是说，我们需要找到此前的一个前缀和模 $p$ 的值为 $target$ 的位置 $j$，使得 $(target + k - cur) \bmod p = 0$。如果找到，我们就可以将 $j + 1$ 到 $i$ 这一段闭区间子数组 $\textit{nums}[j+1,..i]$ 删除，使得剩余元素之和模 $p$ 的值为 $0$。
 
-因此，如果存在一个 $target = (cur - k + p) \bmod p$，那么我们可以更新答案为 $\min(ans, i - j)$。接下来，我们更新 $last[cur]$ 的值为 $i$。继续遍历数组 $nums$，直到遍历结束，即可得到答案。
+因此，如果存在一个 $target = (cur - k + p) \bmod p$，那么我们可以更新答案为 $\min(ans, i - j)$。接下来，我们更新 $last[cur]$ 的值为 $i$。继续遍历数组 $\textit{nums}$，直到遍历结束，即可得到答案。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -234,6 +234,79 @@ function minSubarray(nums: number[], p: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn min_subarray(nums: Vec<i32>, p: i32) -> i32 {
+        let mut k = 0;
+        for &x in &nums {
+            k = (k + x) % p;
+        }
+        if k == 0 {
+            return 0;
+        }
+
+        let mut last = HashMap::new();
+        last.insert(0, -1);
+        let n = nums.len();
+        let mut ans = n as i32;
+        let mut cur = 0;
+
+        for i in 0..n {
+            cur = (cur + nums[i]) % p;
+            let target = (cur - k + p) % p;
+            if let Some(&prev_idx) = last.get(&target) {
+                ans = ans.min(i as i32 - prev_idx);
+            }
+            last.insert(cur, i as i32);
+        }
+
+        if ans == n as i32 {
+            -1
+        } else {
+            ans
+        }
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+var minSubarray = function (nums, p) {
+    let k = 0;
+    for (const x of nums) {
+        k = (k + x) % p;
+    }
+    if (k === 0) {
+        return 0;
+    }
+    const last = new Map();
+    last.set(0, -1);
+    const n = nums.length;
+    let ans = n;
+    let cur = 0;
+    for (let i = 0; i < n; ++i) {
+        cur = (cur + nums[i]) % p;
+        const target = (cur - k + p) % p;
+        if (last.has(target)) {
+            const j = last.get(target);
+            ans = Math.min(ans, i - j);
+        }
+        last.set(cur, i);
+    }
+    return ans === n ? -1 : ans;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1500-1599/1590.Make Sum Divisible by P/README_EN.md

@@ -66,7 +66,19 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Prefix Sum + Hash Table
+
+First, we calculate the sum of all elements in the array $\textit{nums}$ modulo $p$, denoted as $k$. If $k$ is $0$, it means the sum of all elements in the array $\textit{nums}$ is a multiple of $p$, so we directly return $0$.
+
+If $k$ is not $0$, we need to find the shortest subarray such that removing this subarray makes the sum of the remaining elements modulo $p$ equal to $0$.
+
+We can traverse the array $\textit{nums}$, maintaining the current prefix sum modulo $p$, denoted as $cur$. We use a hash table $last$ to record the last occurrence of each prefix sum modulo $p$.
+
+If there exists a subarray ending at $\textit{nums}[i]$ such that removing this subarray makes the sum of the remaining elements modulo $p$ equal to $0$, we need to find a previous prefix sum modulo $p$ equal to $target$ at position $j$ such that $(target + k - cur) \bmod p = 0$. If found, we can remove the subarray $\textit{nums}[j+1,..i]$ to make the sum of the remaining elements modulo $p$ equal to $0$.
+
+Therefore, if there exists a $target = (cur - k + p) \bmod p$, we can update the answer to $\min(ans, i - j)$. Then, we update $last[cur]$ to $i$. We continue traversing the array $\textit{nums}$ until the end to get the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
@@ -210,6 +222,79 @@ function minSubarray(nums: number[], p: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn min_subarray(nums: Vec<i32>, p: i32) -> i32 {
+        let mut k = 0;
+        for &x in &nums {
+            k = (k + x) % p;
+        }
+        if k == 0 {
+            return 0;
+        }
+
+        let mut last = HashMap::new();
+        last.insert(0, -1);
+        let n = nums.len();
+        let mut ans = n as i32;
+        let mut cur = 0;
+
+        for i in 0..n {
+            cur = (cur + nums[i]) % p;
+            let target = (cur - k + p) % p;
+            if let Some(&prev_idx) = last.get(&target) {
+                ans = ans.min(i as i32 - prev_idx);
+            }
+            last.insert(cur, i as i32);
+        }
+
+        if ans == n as i32 {
+            -1
+        } else {
+            ans
+        }
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+var minSubarray = function (nums, p) {
+    let k = 0;
+    for (const x of nums) {
+        k = (k + x) % p;
+    }
+    if (k === 0) {
+        return 0;
+    }
+    const last = new Map();
+    last.set(0, -1);
+    const n = nums.length;
+    let ans = n;
+    let cur = 0;
+    for (let i = 0; i < n; ++i) {
+        cur = (cur + nums[i]) % p;
+        const target = (cur - k + p) % p;
+        if (last.has(target)) {
+            const j = last.get(target);
+            ans = Math.min(ans, i - j);
+        }
+        last.set(cur, i);
+    }
+    return ans === n ? -1 : ans;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1500-1599/1590.Make Sum Divisible by P/Solution.js

@@ -0,0 +1,29 @@
+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+var minSubarray = function (nums, p) {
+    let k = 0;
+    for (const x of nums) {
+        k = (k + x) % p;
+    }
+    if (k === 0) {
+        return 0;
+    }
+    const last = new Map();
+    last.set(0, -1);
+    const n = nums.length;
+    let ans = n;
+    let cur = 0;
+    for (let i = 0; i < n; ++i) {
+        cur = (cur + nums[i]) % p;
+        const target = (cur - k + p) % p;
+        if (last.has(target)) {
+            const j = last.get(target);
+            ans = Math.min(ans, i - j);
+        }
+        last.set(cur, i);
+    }
+    return ans === n ? -1 : ans;
+};

solution/1500-1599/1590.Make Sum Divisible by P/Solution.rs

@@ -0,0 +1,34 @@
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn min_subarray(nums: Vec<i32>, p: i32) -> i32 {
+        let mut k = 0;
+        for &x in &nums {
+            k = (k + x) % p;
+        }
+        if k == 0 {
+            return 0;
+        }
+
+        let mut last = HashMap::new();
+        last.insert(0, -1);
+        let n = nums.len();
+        let mut ans = n as i32;
+        let mut cur = 0;
+
+        for i in 0..n {
+            cur = (cur + nums[i]) % p;
+            let target = (cur - k + p) % p;
+            if let Some(&prev_idx) = last.get(&target) {
+                ans = ans.min(i as i32 - prev_idx);
+            }
+            last.insert(cur, i as i32);
+        }
+
+        if ans == n as i32 {
+            -1
+        } else {
+            ans
+        }
+    }
+}