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Mar 6, 2025
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feat: add solutions to lc problem: No.0985 #4131

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74 changes: 55 additions & 19 deletions solution/0900-0999/0985.Sum of Even Numbers After Queries/README.md
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Expand Up @@ -59,15 +59,11 @@ tags:

### 方法一：模拟

我们用一个变量 $s$ 记录数组 $nums$ 中所有偶数的和。
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。

对于每次查询 $(v, i)$：
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 $\textit{nums}[i]$，然后将 $\textit{s}$ 加入答案数组。

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 $v$；

若 $nums[i]$ 为偶数，则将 $s$ 加上 $nums[i]$，然后将 $s$ 加入答案数组。

时间复杂度 $O(n + m)$，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 $O(1)$。
时间复杂度 $O(n + m)$，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 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -178,12 +174,7 @@ func sumEvenAfterQueries(nums []int, queries [][]int) (ans []int) {

```ts
function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
let s = 0;
for (const x of nums) {
if (x % 2 === 0) {
s += x;
}
}
let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
const ans: number[] = [];
for (const [v, i] of queries) {
if (nums[i] % 2 === 0) {
Expand All @@ -199,6 +190,31 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
}
```

#### Rust

```rust
impl Solution {
pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
let mut ans = Vec::with_capacity(queries.len());

for query in queries {
let (v, i) = (query[0], query[1] as usize);
if nums[i] % 2 == 0 {
s -= nums[i];
}
nums[i] += v;
if nums[i] % 2 == 0 {
s += nums[i];
}
ans.push(s);
}

ans
}
}
```

#### JavaScript

```js
Expand All @@ -208,12 +224,7 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
* @return {number[]}
*/
var sumEvenAfterQueries = function (nums, queries) {
let s = 0;
for (const x of nums) {
if (x % 2 === 0) {
s += x;
}
}
let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
const ans = [];
for (const [v, i] of queries) {
if (nums[i] % 2 === 0) {
Expand All @@ -229,6 +240,31 @@ var sumEvenAfterQueries = function (nums, queries) {
};
```

#### C#

```cs
public class Solution {
public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
int s = nums.Where(x => x % 2 == 0).Sum();
int[] ans = new int[queries.Length];

for (int j = 0; j < queries.Length; j++) {
int v = queries[j][0], i = queries[j][1];
if (nums[i] % 2 == 0) {
s -= nums[i];
}
nums[i] += v;
if (nums[i] % 2 == 0) {
s += nums[i];
}
ans[j] = s;
}

return ans;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
72 changes: 59 additions & 13 deletions solution/0900-0999/0985.Sum of Even Numbers After Queries/README_EN.md
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Expand Up @@ -60,7 +60,13 @@ After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values

<!-- solution:start -->

### Solution 1
### Solution 1: Simulation

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 $\textit{nums}$.

For each query $(v, i)$, 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 $\textit{s}$, and then add $\textit{s}$ to the answer array.

The time complexity is $O(n + m)$, where $n$ and $m$ are the lengths of the arrays $\textit{nums}$ and $\textit{queries}$, respectively. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -171,12 +177,7 @@ func sumEvenAfterQueries(nums []int, queries [][]int) (ans []int) {

```ts
function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
let s = 0;
for (const x of nums) {
if (x % 2 === 0) {
s += x;
}
}
let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
const ans: number[] = [];
for (const [v, i] of queries) {
if (nums[i] % 2 === 0) {
Expand All @@ -192,6 +193,31 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
}
```

#### Rust

```rust
impl Solution {
pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
let mut ans = Vec::with_capacity(queries.len());

for query in queries {
let (v, i) = (query[0], query[1] as usize);
if nums[i] % 2 == 0 {
s -= nums[i];
}
nums[i] += v;
if nums[i] % 2 == 0 {
s += nums[i];
}
ans.push(s);
}

ans
}
}
```

#### JavaScript

```js
Expand All @@ -201,12 +227,7 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
* @return {number[]}
*/
var sumEvenAfterQueries = function (nums, queries) {
let s = 0;
for (const x of nums) {
if (x % 2 === 0) {
s += x;
}
}
let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
const ans = [];
for (const [v, i] of queries) {
if (nums[i] % 2 === 0) {
Expand All @@ -222,6 +243,31 @@ var sumEvenAfterQueries = function (nums, queries) {
};
```

#### C#

```cs
public class Solution {
public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
int s = nums.Where(x => x % 2 == 0).Sum();
int[] ans = new int[queries.Length];

for (int j = 0; j < queries.Length; j++) {
int v = queries[j][0], i = queries[j][1];
if (nums[i] % 2 == 0) {
s -= nums[i];
}
nums[i] += v;
if (nums[i] % 2 == 0) {
s += nums[i];
}
ans[j] = s;
}

return ans;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
20 changes: 20 additions & 0 deletions solution/0900-0999/0985.Sum of Even Numbers After Queries/Solution.cs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
public class Solution {
public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
int s = nums.Where(x => x % 2 == 0).Sum();
int[] ans = new int[queries.Length];

for (int j = 0; j < queries.Length; j++) {
int v = queries[j][0], i = queries[j][1];
if (nums[i] % 2 == 0) {
s -= nums[i];
}
nums[i] += v;
if (nums[i] % 2 == 0) {
s += nums[i];
}
ans[j] = s;
}

return ans;
}
}
7 changes: 1 addition & 6 deletions solution/0900-0999/0985.Sum of Even Numbers After Queries/Solution.js
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Original file line number Diff line number Diff line change
Expand Up @@ -4,12 +4,7 @@
* @return {number[]}
*/
var sumEvenAfterQueries = function (nums, queries) {
let s = 0;
for (const x of nums) {
if (x % 2 === 0) {
s += x;
}
}
let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
const ans = [];
for (const [v, i] of queries) {
if (nums[i] % 2 === 0) {
Expand Down
20 changes: 20 additions & 0 deletions solution/0900-0999/0985.Sum of Even Numbers After Queries/Solution.rs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
impl Solution {
pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
let mut ans = Vec::with_capacity(queries.len());

for query in queries {
let (v, i) = (query[0], query[1] as usize);
if nums[i] % 2 == 0 {
s -= nums[i];
}
nums[i] += v;
if nums[i] % 2 == 0 {
s += nums[i];
}
ans.push(s);
}

ans
}
}
7 changes: 1 addition & 6 deletions solution/0900-0999/0985.Sum of Even Numbers After Queries/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,10 +1,5 @@
function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
let s = 0;
for (const x of nums) {
if (x % 2 === 0) {
s += x;
}
}
let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
const ans: number[] = [];
for (const [v, i] of queries) {
if (nums[i] % 2 === 0) {
Expand Down