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Commit f65ceab

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feat: add solutions to lc problem: No.0985 (#4131)
No.0985.Sum of Even Numbers After Queries
1 parent aa1a162 commit f65ceab

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‎solution/0900-0999/0985.Sum of Even Numbers After Queries/README.md‎

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### 方法一:模拟
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我们用一个变量 $s$ 记录数组 $nums$ 中所有偶数的和。
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我们用一个整型变量 $\textit{s}$ 记录数组 $\textit{nums}$ 中所有偶数的和,初始时 $\textit{s}$ 为数组 $\textit{nums}$ 中所有偶数的和。
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对于每次查询 $(v, i)$:
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对于每次查询 $(v, i)$,我们首先判断 $\textit{nums}[i]$ 是否为偶数,若 $\textit{nums}[i]$ 为偶数,则将 $\textit{s}$ 减去 $\textit{nums}[i]$;然后将 $\textit{nums}[i]$ 加上 $v$;若 $\textit{nums}[i]$ 为偶数,则将 $\textit{s}$ 加上 $\textit{nums}[i],ドル然后将 $\textit{s}$ 加入答案数组。
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我们先判断 $nums[i]$ 是否为偶数,若 $nums[i]$ 为偶数,则将 $s$ 减去 $nums[i]$;然后将 $nums[i]$ 加上 $v$;
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若 $nums[i]$ 为偶数,则将 $s$ 加上 $nums[i],ドル然后将 $s$ 加入答案数组。
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时间复杂度 $O(n + m),ドル其中 $n$ 和 $m$ 分别为数组 $nums$ 和 $queries$ 的长度。忽略答案数组的空间消耗,空间复杂度 $O(1)$。
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时间复杂度 $O(n + m),ドル其中 $n$ 和 $m$ 分别为数组 $\textit{nums}$ 和 $\textit{queries}$ 的长度。忽略答案数组的空间消耗,空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -178,12 +174,7 @@ func sumEvenAfterQueries(nums []int, queries [][]int) (ans []int) {
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```ts
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function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
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const ans: number[] = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
@@ -199,6 +190,31 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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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 sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
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let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
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let mut ans = Vec::with_capacity(queries.len());
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for query in queries {
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let (v, i) = (query[0], query[1] as usize);
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if nums[i] % 2 == 0 {
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s -= nums[i];
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}
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nums[i] += v;
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if nums[i] % 2 == 0 {
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s += nums[i];
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}
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ans.push(s);
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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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#### JavaScript
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```js
@@ -208,12 +224,7 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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* @return {number[]}
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*/
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var sumEvenAfterQueries = function (nums, queries) {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
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const ans = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
@@ -229,6 +240,31 @@ var sumEvenAfterQueries = function (nums, queries) {
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};
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```
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#### C#
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```cs
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public class Solution {
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public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
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int s = nums.Where(x => x % 2 == 0).Sum();
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int[] ans = new int[queries.Length];
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for (int j = 0; j < queries.Length; j++) {
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int v = queries[j][0], i = queries[j][1];
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if (nums[i] % 2 == 0) {
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s -= nums[i];
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}
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nums[i] += v;
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if (nums[i] % 2 == 0) {
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s += nums[i];
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}
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ans[j] = s;
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}
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return 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/0900-0999/0985.Sum of Even Numbers After Queries/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -60,7 +60,13 @@ After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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We use an integer variable $\textit{s}$ to record the sum of all even numbers in the array $\textit{nums}$. Initially, $\textit{s}$ is the sum of all even numbers in the array $\textit{nums}$.
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For each query $(v, i),ドル we first check if $\textit{nums}[i]$ is even. If $\textit{nums}[i]$ is even, we subtract $\textit{nums}[i]$ from $\textit{s}$. Then, we add $v$ to $\textit{nums}[i]$. If $\textit{nums}[i]$ is even, we add $\textit{nums}[i]$ to $\textit{s},ドル and then add $\textit{s}$ to the answer array.
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The time complexity is $O(n + m),ドル where $n$ and $m$ are the lengths of the arrays $\textit{nums}$ and $\textit{queries},ドル respectively. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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<!-- tabs:start -->
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```ts
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function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
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const ans: number[] = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
@@ -192,6 +193,31 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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}
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```
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#### Rust
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```rust
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impl Solution {
200+
pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
201+
let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
202+
let mut ans = Vec::with_capacity(queries.len());
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204+
for query in queries {
205+
let (v, i) = (query[0], query[1] as usize);
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if nums[i] % 2 == 0 {
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s -= nums[i];
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}
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nums[i] += v;
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if nums[i] % 2 == 0 {
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s += nums[i];
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}
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ans.push(s);
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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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#### JavaScript
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```js
@@ -201,12 +227,7 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
201227
* @return {number[]}
202228
*/
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var sumEvenAfterQueries = function (nums, queries) {
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let s = 0;
205-
for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
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const ans = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
@@ -222,6 +243,31 @@ var sumEvenAfterQueries = function (nums, queries) {
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};
223244
```
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#### C#
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```cs
249+
public class Solution {
250+
public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
251+
int s = nums.Where(x => x % 2 == 0).Sum();
252+
int[] ans = new int[queries.Length];
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for (int j = 0; j < queries.Length; j++) {
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int v = queries[j][0], i = queries[j][1];
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if (nums[i] % 2 == 0) {
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s -= nums[i];
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}
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nums[i] += v;
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if (nums[i] % 2 == 0) {
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s += nums[i];
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}
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ans[j] = s;
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}
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return 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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public class Solution {
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public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
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int s = nums.Where(x => x % 2 == 0).Sum();
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int[] ans = new int[queries.Length];
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6+
for (int j = 0; j < queries.Length; j++) {
7+
int v = queries[j][0], i = queries[j][1];
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if (nums[i] % 2 == 0) {
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s -= nums[i];
10+
}
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nums[i] += v;
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if (nums[i] % 2 == 0) {
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s += nums[i];
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}
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ans[j] = s;
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}
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return ans;
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}
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}

‎solution/0900-0999/0985.Sum of Even Numbers After Queries/Solution.js‎

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Original file line numberDiff line numberDiff line change
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* @return {number[]}
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*/
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var sumEvenAfterQueries = function (nums, queries) {
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let s = 0;
8-
for (const x of nums) {
9-
if (x % 2 === 0) {
10-
s += x;
11-
}
12-
}
7+
let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
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const ans = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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Original file line numberDiff line numberDiff line change
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impl Solution {
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pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
3+
let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
4+
let mut ans = Vec::with_capacity(queries.len());
5+
6+
for query in queries {
7+
let (v, i) = (query[0], query[1] as usize);
8+
if nums[i] % 2 == 0 {
9+
s -= nums[i];
10+
}
11+
nums[i] += v;
12+
if nums[i] % 2 == 0 {
13+
s += nums[i];
14+
}
15+
ans.push(s);
16+
}
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ans
19+
}
20+
}

‎solution/0900-0999/0985.Sum of Even Numbers After Queries/Solution.ts‎

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function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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let s = 0;
3-
for (const x of nums) {
4-
if (x % 2 === 0) {
5-
s += x;
6-
}
7-
}
2+
let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
83
const ans: number[] = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {

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