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feat: add solutions to lc problem: No.2389 #4027

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feat: add solutions to lc problem: No.2389 #4027

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112 changes: 45 additions & 67 deletions solution/2300-2399/2389.Longest Subsequence With Limited Sum/README.md
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Expand Up @@ -67,11 +67,11 @@ tags:

### 方法一：排序 + 前缀和 + 二分查找

根据题目描述，对于每个 $queries[i]$，我们需要找到一个子序列，使得该子序列的元素和不超过 $queries[i]$，且该子序列的长度最大化。显然，我们应该选择尽可能小的元素，这样才能使得子序列的长度最大化。
根据题目描述，对于每个 $\textit{queries[i]}$，我们需要找到一个子序列，使得该子序列的元素和不超过 $\textit{queries[i]}$，且该子序列的长度最大化。显然，我们应该选择尽可能小的元素，这样才能使得子序列的长度最大化。

因此，我们可以先将数组 $nums$ 进行升序排序，然后对于每个 $queries[i]$，我们可以使用二分查找，找到最小的下标 $j$，使得 $nums[0] + nums[1] + \cdots + nums[j] \gt queries[i]$。此时 $nums[0] + nums[1] + \cdots + nums[j - 1]$ 就是满足条件的子序列的元素和，且该子序列的长度为 $j$。因此，我们可以将 $j$ 加入答案数组中。
因此，我们可以先将数组 $\textit{nums}$ 进行升序排序，然后对于每个 $\textit{queries[i]}$，我们可以使用二分查找，找到最小的下标 $j$，使得 $\textit{nums}[0] + \textit{nums}[1] + \cdots + \textit{nums}[j] > \textit{queries[i]}$。此时 $\textit{nums}[0] + \textit{nums}[1] + \cdots + \textit{nums}[j - 1]$ 就是满足条件的子序列的元素和，且该子序列的长度为 $j$。因此，我们可以将 $j$ 加入答案数组中。

时间复杂度 $O((n + m) \times \log n)$，空间复杂度 $O(n)$ 或 $O(\log n)$。其中 $n$ 和 $m$ 分别是数组 $nums$ 和 $queries$ 的长度。
时间复杂度 $O((n + m) \times \log n)$，空间复杂度 $O(n)$ 或 $O(\log n)$。其中 $n$ 和 $m$ 分别是数组 $\textit{nums}$ 和 $\textit{queries}$ 的长度。

<!-- tabs:start -->

Expand All @@ -97,23 +97,11 @@ class Solution {
int m = queries.length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
ans[i] = search(nums, queries[i]);
int j = Arrays.binarySearch(nums, queries[i] + 1);
ans[i] = j < 0 ? -j - 1 : j;
}
return ans;
}

private int search(int[] nums, int x) {
int l = 0, r = nums.length;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}
```

Expand All @@ -123,13 +111,13 @@ class Solution {
class Solution {
public:
vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
for (int i = 1; i < nums.size(); i++) {
nums[i] += nums[i - 1];
}
vector<int> ans;
for (auto& q : queries) {
ans.push_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
for (const auto& q : queries) {
ans.emplace_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
}
return ans;
}
Expand Down Expand Up @@ -159,24 +147,7 @@ function answerQueries(nums: number[], queries: number[]): number[] {
for (let i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
}
const ans: number[] = [];
const search = (nums: number[], x: number) => {
let l = 0;
let r = nums.length;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
for (const q of queries) {
ans.push(search(nums, q));
}
return ans;
return queries.map(q => _.sortedIndex(nums, q + 1));
}
```

Expand All @@ -185,48 +156,55 @@ function answerQueries(nums: number[], queries: number[]): number[] {
```rust
impl Solution {
pub fn answer_queries(mut nums: Vec<i32>, queries: Vec<i32>) -> Vec<i32> {
let n = nums.len();
nums.sort();
queries
.into_iter()
.map(|query| {
let mut sum = 0;
for i in 0..n {
sum += nums[i];
if sum > query {
return i as i32;
}
}
n as i32
})
.collect()

for i in 1..nums.len() {
nums[i] += nums[i - 1];
}

queries.iter().map(|&q| {
match nums.binary_search(&q) {
Ok(idx) => idx as i32 + 1,
Err(idx) => idx as i32,
}
}).collect()
}
}
```

#### JavaScript

```js
/**
* @param {number[]} nums
* @param {number[]} queries
* @return {number[]}
*/
var answerQueries = function (nums, queries) {
nums.sort((a, b) => a - b);
for (let i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
}
return queries.map(q => _.sortedIndex(nums, q + 1));
};
```

#### C#

```cs
public class Solution {
public int[] AnswerQueries(int[] nums, int[] queries) {
int[] result = new int[queries.Length];
Array.Sort(nums);
for (int i = 0; i < queries.Length; i++) {
result[i] = getSubsequent(nums, queries[i]);
for (int i = 1; i < nums.Length; ++i) {
nums[i] += nums[i - 1];
}
return result;

}

public int getSubsequent(int[] nums,int query) {
int sum = 0;
for (int i = 0; i < nums.Length; i++) {
sum += nums[i];
if (sum > query) {
return i;
}
int m = queries.Length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
int j = Array.BinarySearch(nums, queries[i] + 1);
ans[i] = j < 0 ? -j - 1 : j;
}
return nums.Length;
return ans;
}
}
```
Expand Down
112 changes: 45 additions & 67 deletions solution/2300-2399/2389.Longest Subsequence With Limited Sum/README_EN.md
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Expand Up @@ -65,11 +65,11 @@ tags:

### Solution 1: Sorting + Prefix Sum + Binary Search

According to the problem description, for each $queries[i]$, we need to find a subsequence such that the sum of its elements does not exceed $queries[i]$ and the length of this subsequence is maximized. Obviously, we should choose the smallest possible elements to maximize the length of the subsequence.
According to the problem description, for each $\textit{queries[i]}$, we need to find a subsequence such that the sum of its elements does not exceed $\textit{queries[i]}$ and the length of the subsequence is maximized. Obviously, we should choose the smallest possible elements to maximize the length of the subsequence.

Therefore, we can first sort the array $nums$ in ascending order. Then, for each $queries[i]$, we can use binary search to find the smallest index $j$ such that $nums[0] + nums[1] + \cdots + nums[j] \gt queries[i]$. At this point, $nums[0] + nums[1] + \cdots + nums[j - 1]$ is the sum of the elements of the subsequence that meets the condition, and the length of this subsequence is $j$. Therefore, we can add $j$ to the answer array.
Therefore, we can first sort the array $\textit{nums}$ in ascending order, and then for each $\textit{queries[i]}$, we can use binary search to find the smallest index $j$ such that $\textit{nums}[0] + \textit{nums}[1] + \cdots + \textit{nums}[j] > \textit{queries[i]}$. At this point, $\textit{nums}[0] + \textit{nums}[1] + \cdots + \textit{nums}[j - 1]$ is the sum of the elements of the subsequence that meets the condition, and the length of this subsequence is $j$. Therefore, we can add $j$ to the answer array.

The time complexity is $O((n + m) \times \log n)$, and the space complexity is $O(n)$ or $O(\log n)$. Here, $n$ and $m$ are the lengths of the arrays $nums$ and $queries$, respectively.
The time complexity is $O((n + m) \times \log n)$, and the space complexity is $O(n)$ or $O(\log n)$. Here, $n$ and $m$ are the lengths of the arrays $\textit{nums}$ and $\textit{queries}$, respectively.

<!-- tabs:start -->

Expand All @@ -95,23 +95,11 @@ class Solution {
int m = queries.length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
ans[i] = search(nums, queries[i]);
int j = Arrays.binarySearch(nums, queries[i] + 1);
ans[i] = j < 0 ? -j - 1 : j;
}
return ans;
}

private int search(int[] nums, int x) {
int l = 0, r = nums.length;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}
```

Expand All @@ -121,13 +109,13 @@ class Solution {
class Solution {
public:
vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
for (int i = 1; i < nums.size(); i++) {
nums[i] += nums[i - 1];
}
vector<int> ans;
for (auto& q : queries) {
ans.push_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
for (const auto& q : queries) {
ans.emplace_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
}
return ans;
}
Expand Down Expand Up @@ -157,24 +145,7 @@ function answerQueries(nums: number[], queries: number[]): number[] {
for (let i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
}
const ans: number[] = [];
const search = (nums: number[], x: number) => {
let l = 0;
let r = nums.length;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
for (const q of queries) {
ans.push(search(nums, q));
}
return ans;
return queries.map(q => _.sortedIndex(nums, q + 1));
}
```

Expand All @@ -183,48 +154,55 @@ function answerQueries(nums: number[], queries: number[]): number[] {
```rust
impl Solution {
pub fn answer_queries(mut nums: Vec<i32>, queries: Vec<i32>) -> Vec<i32> {
let n = nums.len();
nums.sort();
queries
.into_iter()
.map(|query| {
let mut sum = 0;
for i in 0..n {
sum += nums[i];
if sum > query {
return i as i32;
}
}
n as i32
})
.collect()

for i in 1..nums.len() {
nums[i] += nums[i - 1];
}

queries.iter().map(|&q| {
match nums.binary_search(&q) {
Ok(idx) => idx as i32 + 1,
Err(idx) => idx as i32,
}
}).collect()
}
}
```

#### JavaScript

```js
/**
* @param {number[]} nums
* @param {number[]} queries
* @return {number[]}
*/
var answerQueries = function (nums, queries) {
nums.sort((a, b) => a - b);
for (let i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
}
return queries.map(q => _.sortedIndex(nums, q + 1));
};
```

#### C#

```cs
public class Solution {
public int[] AnswerQueries(int[] nums, int[] queries) {
int[] result = new int[queries.Length];
Array.Sort(nums);
for (int i = 0; i < queries.Length; i++) {
result[i] = getSubsequent(nums, queries[i]);
for (int i = 1; i < nums.Length; ++i) {
nums[i] += nums[i - 1];
}
return result;

}

public int getSubsequent(int[] nums,int query) {
int sum = 0;
for (int i = 0; i < nums.Length; i++) {
sum += nums[i];
if (sum > query) {
return i;
}
int m = queries.Length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
int j = Array.BinarySearch(nums, queries[i] + 1);
ans[i] = j < 0 ? -j - 1 : j;
}
return nums.Length;
return ans;
}
}
```
Expand Down
8 changes: 4 additions & 4 deletions solution/2300-2399/2389.Longest Subsequence With Limited Sum/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,14 +1,14 @@
class Solution {
public:
vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
for (int i = 1; i < nums.size(); i++) {
nums[i] += nums[i - 1];
}
vector<int> ans;
for (auto& q : queries) {
ans.push_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
for (const auto& q : queries) {
ans.emplace_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
}
return ans;
}
};
};
23 changes: 8 additions & 15 deletions solution/2300-2399/2389.Longest Subsequence With Limited Sum/Solution.cs
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,22 +1,15 @@
public class Solution {
public int[] AnswerQueries(int[] nums, int[] queries) {
int[] result = new int[queries.Length];
Array.Sort(nums);
for (int i = 0; i < queries.Length; i++) {
result[i] = getSubsequent(nums, queries[i]);
for (int i = 1; i < nums.Length; ++i) {
nums[i] += nums[i - 1];
}
return result;

}

public int getSubsequent(int[] nums,int query) {
int sum = 0;
for (int i = 0; i < nums.Length; i++) {
sum += nums[i];
if (sum > query) {
return i;
}
int m = queries.Length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
int j = Array.BinarySearch(nums, queries[i] + 1);
ans[i] = j < 0 ? -j - 1 : j;
}
return nums.Length;
return ans;
}
}
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