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Sep 5, 2024
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feat: add solutions to lc problem: No.2576 #3490

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93 changes: 47 additions & 46 deletions solution/2500-2599/2576.Find the Maximum Number of Marked Indices/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -78,9 +78,11 @@ tags:

### 方法一：贪心 + 双指针

为了将下标尽可能多地标记，我们可以将数组 `nums` 排序，然后从左到右遍历数组，对于每个下标 $i$，我们在数组的右半部分找到第一个满足 $2 \times nums[i] \leq nums[j]$ 的下标 $j$，然后标记下标 $i$ 和 $j$。继续遍历下一个下标 $i$。当我们遍历完数组的右半部分时，说明标记已经完成，此时标记的下标数目即为答案。
根据题目描述，题目最多产生 $n / 2$ 组标记，其中 $n$ 为数组 $\textit{nums}$ 的长度。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `nums` 的长度。
为了将下标尽可能多地标记，我们可以将数组 $\textit{nums}$ 排序，接下来，我们遍历右半部分的每个元素 $\textit{nums}[j]$，用一个指针 $\textit{i}$ 指向左半部分的最小元素，如果 $\textit{nums}[i] \times 2 \leq \textit{nums}[j]$，则可以标记下标 $\textit{i}$ 和 $\textit{j}$，我们将 $\textit{i}$ 向右移动一个位置。继续遍历右半部分的元素，直到到达数组的末尾。此时，我们可以标记的下标数目为 $\textit{i} \times 2$。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。

<!-- tabs:start -->

Expand All @@ -90,16 +92,11 @@ tags:
class Solution:
def maxNumOfMarkedIndices(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
i, j = 0, (n + 1) // 2
ans = 0
while j < n:
while j < n and nums[i] * 2 > nums[j]:
j += 1
if j < n:
ans += 2
i, j = i + 1, j + 1
return ans
i, n = 0, len(nums)
for x in nums[(n + 1) // 2 :]:
if nums[i] * 2 <= x:
i += 1
return i * 2
```

#### Java
Expand All @@ -108,17 +105,13 @@ class Solution:
class Solution {
public int maxNumOfMarkedIndices(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int ans = 0;
for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
}
if (j < n) {
ans += 2;
int i = 0, n = nums.length;
for (int j = (n + 1) / 2; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
}
return ans;
return i * 2;
}
}
```
Expand All @@ -129,18 +122,14 @@ class Solution {
class Solution {
public:
int maxNumOfMarkedIndices(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
int ans = 0;
for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
}
if (j < n) {
ans += 2;
ranges::sort(nums);
int i = 0, n = nums.size();
for (int j = (n + 1) / 2; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
}
return ans;
return i * 2;
}
};
```
Expand All @@ -150,16 +139,13 @@ public:
```go
func maxNumOfMarkedIndices(nums []int) (ans int) {
sort.Ints(nums)
n := len(nums)
for i, j := 0, (n+1)/2; j < n; i, j = i+1, j+1 {
for j < n && nums[i]*2 > nums[j] {
j++
}
if j < n {
ans += 2
i, n := 0, len(nums)
for _, x := range nums[(n+1)/2:] {
if nums[i]*2 <= x {
i++
}
}
return
return i * 2
}
```

Expand All @@ -169,16 +155,31 @@ func maxNumOfMarkedIndices(nums []int) (ans int) {
function maxNumOfMarkedIndices(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
let ans = 0;
for (let i = 0, j = Math.floor((n + 1) / 2); j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
let i = 0;
for (let j = (n + 1) >> 1; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
if (j < n) {
ans += 2;
}
return i * 2;
}
```

#### Rust

```rust
impl Solution {
pub fn max_num_of_marked_indices(mut nums: Vec<i32>) -> i32 {
nums.sort();
let mut i = 0;
let n = nums.len();
for j in (n + 1) / 2..n {
if nums[i] * 2 <= nums[j] {
i += 1;
}
}
(i * 2) as i32
}
return ans;
}
```

Expand Down
95 changes: 48 additions & 47 deletions solution/2500-2599/2576.Find the Maximum Number of Marked Indices/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -70,7 +70,7 @@ Since there is no other operation, the answer is 4.
</ul>

<p>&nbsp;</p>
<style type="text/css">.spoilerbutton {display:block; border:dashed; padding: 0px 0px; margin:10px 0px; font-size:150%; font-weight: bold; color:#000000; background-color:cyan; outline:0;
<style type="text/css">.spoilerbutton {display:block; border:dashed; padding: 0px 0px; margin:10px 0px; font-size:150%; font-weight: bold; color:#000000; background-color:cyan; outline:0;
}
.spoiler {overflow:hidden;}
.spoiler > div {-webkit-transition: all 0s ease;-moz-transition: margin 0s ease;-o-transition: all 0s ease;transition: margin 0s ease;}
Expand All @@ -86,9 +86,11 @@ Since there is no other operation, the answer is 4.

### Solution 1: Greedy + Two Pointers

In order to mark as many indices as possible, we can sort the array `nums`, and then traverse the array from left to right. For each index $i$, we find the first index $j$ in the right half of the array that satisfies $2 \times nums[i] \leq nums[j]$, and then mark indices $i$ and $j$. Continue to traverse the next index $i$. When we have traversed the right half of the array, it means that the marking is complete, and the number of marked indices is the answer.
According to the problem description, the problem can generate at most $n / 2$ pairs of indices, where $n$ is the length of the array $\textit{nums}$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array `nums`.
To mark as many indices as possible, we can sort the array $\textit{nums}$. Next, we traverse each element $\textit{nums}[j]$ in the right half of the array, using a pointer $\textit{i}$ to point to the smallest element in the left half. If $\textit{nums}[i] \times 2 \leq \textit{nums}[j]$, we can mark the indices $\textit{i}$ and $\textit{j}$, and move $\textit{i}$ one position to the right. Continue traversing the elements in the right half until reaching the end of the array. At this point, the number of indices we can mark is $\textit{i} \times 2$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.

<!-- tabs:start -->

Expand All @@ -98,16 +100,11 @@ The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log
class Solution:
def maxNumOfMarkedIndices(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
i, j = 0, (n + 1) // 2
ans = 0
while j < n:
while j < n and nums[i] * 2 > nums[j]:
j += 1
if j < n:
ans += 2
i, j = i + 1, j + 1
return ans
i, n = 0, len(nums)
for x in nums[(n + 1) // 2 :]:
if nums[i] * 2 <= x:
i += 1
return i * 2
```

#### Java
Expand All @@ -116,17 +113,13 @@ class Solution:
class Solution {
public int maxNumOfMarkedIndices(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int ans = 0;
for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
}
if (j < n) {
ans += 2;
int i = 0, n = nums.length;
for (int j = (n + 1) / 2; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
}
return ans;
return i * 2;
}
}
```
Expand All @@ -137,18 +130,14 @@ class Solution {
class Solution {
public:
int maxNumOfMarkedIndices(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
int ans = 0;
for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
}
if (j < n) {
ans += 2;
ranges::sort(nums);
int i = 0, n = nums.size();
for (int j = (n + 1) / 2; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
}
return ans;
return i * 2;
}
};
```
Expand All @@ -158,16 +147,13 @@ public:
```go
func maxNumOfMarkedIndices(nums []int) (ans int) {
sort.Ints(nums)
n := len(nums)
for i, j := 0, (n+1)/2; j < n; i, j = i+1, j+1 {
for j < n && nums[i]*2 > nums[j] {
j++
}
if j < n {
ans += 2
i, n := 0, len(nums)
for _, x := range nums[(n+1)/2:] {
if nums[i]*2 <= x {
i++
}
}
return
return i * 2
}
```

Expand All @@ -177,16 +163,31 @@ func maxNumOfMarkedIndices(nums []int) (ans int) {
function maxNumOfMarkedIndices(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
let ans = 0;
for (let i = 0, j = Math.floor((n + 1) / 2); j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
let i = 0;
for (let j = (n + 1) >> 1; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
if (j < n) {
ans += 2;
}
return i * 2;
}
```

#### Rust

```rust
impl Solution {
pub fn max_num_of_marked_indices(mut nums: Vec<i32>) -> i32 {
nums.sort();
let mut i = 0;
let n = nums.len();
for j in (n + 1) / 2..n {
if nums[i] * 2 <= nums[j] {
i += 1;
}
}
(i * 2) as i32
}
return ans;
}
```

Expand Down
18 changes: 7 additions & 11 deletions solution/2500-2599/2576.Find the Maximum Number of Marked Indices/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,17 +1,13 @@
class Solution {
public:
int maxNumOfMarkedIndices(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
int ans = 0;
for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
}
if (j < n) {
ans += 2;
ranges::sort(nums);
int i = 0, n = nums.size();
for (int j = (n + 1) / 2; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
}
return ans;
return i * 2;
}
};
};
15 changes: 6 additions & 9 deletions solution/2500-2599/2576.Find the Maximum Number of Marked Indices/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,10 @@
func maxNumOfMarkedIndices(nums []int) (ans int) {
sort.Ints(nums)
n := len(nums)
for i, j := 0, (n+1)/2; j < n; i, j = i+1, j+1 {
for j < n && nums[i]*2 > nums[j] {
j++
}
if j < n {
ans += 2
i, n := 0, len(nums)
for _, x := range nums[(n+1)/2:] {
if nums[i]*2 <= x {
i++
}
}
return
}
return i * 2
}
16 changes: 6 additions & 10 deletions solution/2500-2599/2576.Find the Maximum Number of Marked Indices/Solution.java
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,16 +1,12 @@
class Solution {
public int maxNumOfMarkedIndices(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int ans = 0;
for (int i = 0, j = (n + 1) / 2; j < n; ++i, ++j) {
while (j < n && nums[i] * 2 > nums[j]) {
++j;
}
if (j < n) {
ans += 2;
int i = 0, n = nums.length;
for (int j = (n + 1) / 2; j < n; ++j) {
if (nums[i] * 2 <= nums[j]) {
++i;
}
}
return ans;
return i * 2;
}
}
}
15 changes: 5 additions & 10 deletions solution/2500-2599/2576.Find the Maximum Number of Marked Indices/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,8 @@
class Solution:
def maxNumOfMarkedIndices(self, nums: List[int]) -> int:
nums.sort()
n = len(nums)
i, j = 0, (n + 1) // 2
ans = 0
while j < n:
while j < n and nums[i] * 2 > nums[j]:
j += 1
if j < n:
ans += 2
i, j = i + 1, j + 1
return ans
i, n = 0, len(nums)
for x in nums[(n + 1) // 2 :]:
if nums[i] * 2 <= x:
i += 1
return i * 2
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