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Commit 4a32c1c

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feat: add solutions to lc problem: No.1524 (#4112)
No.1524.Number of Sub-arrays With Odd Sum
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‎solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README.md‎

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### 方法一:前缀和 + 计数器
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我们定义一个长度为 2ドル$ 的数组 $cnt$ 作为计数器,其中 $cnt[0]$ 和 $cnt[1]$ 分别表示前缀和为偶数和奇数的子数组的个数。初始时 $cnt[0] = 1,ドル而 $cnt[1] = 0$。
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我们定义一个长度为 2ドル$ 的数组 $\textit{cnt}$ 作为计数器,其中 $\textit{cnt}[0]$ 和 $\textit{cnt}[1]$ 分别表示前缀和为偶数和奇数的子数组的个数。初始时 $\textit{cnt}[0] = 1,ドル而 $\textit{cnt}[1] = 0$。
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接下来,我们维护当前的前缀和 $s,ドル初始时 $s = 0$。
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遍历数组 $arr,ドル对于遍历到的每个元素 $x,ドル我们将 $s$ 的值加上 $x,ドル然后根据 $s$ 的奇偶性,将 $cnt[s \mod 2 \oplus 1]$ 的值累加到答案中,然后我们将 $cnt[s \mod 2]$ 的值加 1ドル$。
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遍历数组 $\textit{arr},ドル对于遍历到的每个元素 $x,ドル我们将 $s$ 的值加上 $x,ドル然后根据 $s$ 的奇偶性,将 $\textit{cnt}[s \mod 2 \oplus 1]$ 的值累加到答案中,然后我们将 $\textit{cnt}[s \mod 2]$ 的值加 1ドル$。
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遍历结束后,我们即可得到答案。注意答案的取模运算。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组 $arr$ 的长度。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 $\textit{arr}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
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const MOD: i32 = 1_000_000_007;
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let mut cnt = [1, 0];
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let mut ans = 0;
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let mut s = 0;
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for &x in arr.iter() {
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s += x;
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ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
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cnt[(s & 1) as usize] += 1;
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1500-1599/1524.Number of Sub-arrays With Odd Sum/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Prefix Sum + Counter
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We define an array $\textit{cnt}$ of length 2 as a counter, where $\textit{cnt}[0]$ and $\textit{cnt}[1]$ represent the number of subarrays with even and odd prefix sums, respectively. Initially, $\textit{cnt}[0] = 1$ and $\textit{cnt}[1] = 0$.
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Next, we maintain the current prefix sum $s,ドル initially $s = 0$.
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Traverse the array $\textit{arr},ドル for each element $x$ encountered, add the value of $x$ to $s,ドル then based on the parity of $s,ドル add the value of $\textit{cnt}[s \mod 2 \oplus 1]$ to the answer, and then increment the value of $\textit{cnt}[s \mod 2]$ by 1.
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After the traversal, we get the answer. Note the modulo operation for the answer.
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Time complexity is $O(n),ドル where $n$ is the length of the array $\textit{arr}$. Space complexity is $O(1)$.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
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const MOD: i32 = 1_000_000_007;
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let mut cnt = [1, 0];
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let mut ans = 0;
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let mut s = 0;
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for &x in arr.iter() {
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s += x;
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ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
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cnt[(s & 1) as usize] += 1;
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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impl Solution {
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pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
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const MOD: i32 = 1_000_000_007;
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let mut cnt = [1, 0];
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let mut ans = 0;
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let mut s = 0;
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for &x in arr.iter() {
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s += x;
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ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
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cnt[(s & 1) as usize] += 1;
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}
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ans
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}
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}

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