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feat: add solutions to lc problem: No.3247 (#3379)
No.3247.Number of Subsequences with Odd Sum
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‎solution/2700-2799/2753.Count Houses in a Circular Street II/README.md‎

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comments: true
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difficulty: 困难
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/2700-2799/2753.Count%20Houses%20in%20a%20Circular%20Street%20II/README.md
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tags:
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- Algorithms
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‎solution/2700-2799/2753.Count Houses in a Circular Street II/README_EN.md‎

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comments: true
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difficulty: Hard
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/2700-2799/2753.Count%20Houses%20in%20a%20Circular%20Street%20II/README_EN.md
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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3247.Number%20of%20Subsequences%20with%20Odd%20Sum/README.md
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---
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<!-- problem:start -->
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# [3247. Number of Subsequences with Odd Sum 🔒](https://leetcode.cn/problems/number-of-subsequences-with-odd-sum)
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[English Version](/solution/3200-3299/3247.Number%20of%20Subsequences%20with%20Odd%20Sum/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Given an array <code>nums</code>, return the number of <span data-keyword="subsequence-array">subsequences</span> with an odd sum of elements.</p>
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<p>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,1,1]</span></p>
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<p><strong>Output:</strong> <span class="example-io">4</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The odd-sum subsequences are: <code>[<u><strong>1</strong></u>, 1, 1]</code>, <code>[1, <u><strong>1</strong></u>, 1],</code> <code>[1, 1, <u><strong>1</strong></u>]</code>, <code>[<u><strong>1, 1, 1</strong></u>]</code>.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,2,2]</span></p>
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<p><strong>Output:</strong> <span class="example-io">4</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The odd-sum subsequences are: <code>[<u><strong>1</strong></u>, 2, 2]</code>, <code>[<u><strong>1, 2</strong></u>, 2],</code> <code>[<u><strong>1</strong></u>, 2, <b><u>2</u></b>]</code>, <code>[<u><strong>1, 2, 2</strong></u>]</code>.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.lnegth &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:动态规划
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我们定义 $f[0]$ 表示目前为止的子序列中,和为偶数的子序列个数,而 $f[1]$ 表示目前为止的子序列中,和为奇数的子序列个数。初始时 $f[0] = 0,ドル $f[1] = 0$。
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遍历数组 $\textit{nums},ドル对于每个数 $x$:
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如果 $x$ 为奇数,那么 $f[0]$ 和 $f[1]$ 的更新方式为:
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$$
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\begin{aligned}
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f[0] & = (f[0] + f[1]) \bmod 10^9 + 7, \\
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f[1] & = (f[0] + f[1] + 1) \bmod 10^9 + 7.
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\end{aligned}
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$$
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即,当前的和为偶数的子序列个数等于之前的和为偶数的子序列个数,加上之前的和为奇数的子序列拼上当前数 $x$ 的子序列个数;当前的和为奇数的子序列个数等于之前的和为偶数的子序列拼上当前数 $x$ 的子序列个数,加上之前的和为奇数的子序列个数,再加上一个只包含当前数 $x$ 的子序列。
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如果 $x$ 为偶数,那么 $f[0]$ 和 $f[1]$ 的更新方式为:
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$$
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\begin{aligned}
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f[0] & = (f[0] + f[0] + 1) \bmod 10^9 + 7, \\
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f[1] & = (f[1] + f[1]) \bmod 10^9 + 7.
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\end{aligned}
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$$
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即,当前的和为偶数的子序列个数等于之前的和为偶数的子序列个数,加上之前的和为偶数的子序列拼上当前数 $x$ 的子序列个数,再加上一个只包含当前数 $x$ 的子序列;当前的和为奇数的子序列个数等于之前的和为奇数的子序列拼上当前数 $x$ 的子序列个数,加上之前的和为奇数的子序列个数。
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最终,返回 $f[1]$ 即可。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def subsequenceCount(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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f = [0] * 2
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for x in nums:
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if x % 2:
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f[0], f[1] = (f[0] + f[1]) % mod, (f[0] + f[1] + 1) % mod
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else:
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f[0], f[1] = (f[0] + f[0] + 1) % mod, (f[1] + f[1]) % mod
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return f[1]
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```
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#### Java
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```java
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class Solution {
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public int subsequenceCount(int[] nums) {
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final int mod = (int) 1e9 + 7;
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int[] f = new int[2];
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for (int x : nums) {
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int[] g = new int[2];
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if (x % 2 == 1) {
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g[0] = (f[0] + f[1]) % mod;
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g[1] = (f[0] + f[1] + 1) % mod;
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} else {
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g[0] = (f[0] + f[0] + 1) % mod;
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g[1] = (f[1] + f[1]) % mod;
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}
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f = g;
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}
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return f[1];
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int subsequenceCount(vector<int>& nums) {
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const int mod = 1e9 + 7;
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vector<int> f(2);
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for (int x : nums) {
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vector<int> g(2);
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if (x % 2 == 1) {
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g[0] = (f[0] + f[1]) % mod;
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g[1] = (f[0] + f[1] + 1) % mod;
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} else {
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g[0] = (f[0] + f[0] + 1) % mod;
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g[1] = (f[1] + f[1]) % mod;
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}
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f = g;
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}
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return f[1];
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}
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};
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```
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#### Go
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```go
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func subsequenceCount(nums []int) int {
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mod := int(1e9 + 7)
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f := [2]int{}
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for _, x := range nums {
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g := [2]int{}
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if x%2 == 1 {
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g[0] = (f[0] + f[1]) % mod
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g[1] = (f[0] + f[1] + 1) % mod
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} else {
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g[0] = (f[0] + f[0] + 1) % mod
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g[1] = (f[1] + f[1]) % mod
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}
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f = g
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}
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return f[1]
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}
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```
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#### TypeScript
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```ts
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function subsequenceCount(nums: number[]): number {
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const mod = 1e9 + 7;
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let f = [0, 0];
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for (const x of nums) {
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const g = [0, 0];
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if (x % 2 === 1) {
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g[0] = (f[0] + f[1]) % mod;
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g[1] = (f[0] + f[1] + 1) % mod;
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} else {
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g[0] = (f[0] + f[0] + 1) % mod;
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g[1] = (f[1] + f[1]) % mod;
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}
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f = g;
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}
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return f[1];
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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