Thanks to visit codestin.com
Credit goes to github.com

Skip to content

Sri Hari: Batch-5/Neetcode-ALL/Added-articles #3829

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 4 commits into from
Jan 27, 2025
Merged
Show file tree
Hide file tree
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
338 changes: 338 additions & 0 deletions articles/array-with-elements-not-equal-to-average-of-neighbors.md
Original file line number Diff line number Diff line change
@@ -0,0 +1,338 @@
## 1. Greedy

::tabs-start

```python
class Solution:
def rearrangeArray(self, nums: List[int]) -> List[int]:
nums.sort()
res = []
l, r = 0, len(nums) - 1
while len(res) != len(nums):
res.append(nums[l])
l += 1
if l <= r:
res.append(nums[r])
r -= 1
return res
```

```java
public class Solution {
public int[] rearrangeArray(int[] nums) {
Arrays.sort(nums);
int[] res = new int[nums.length];
int l = 0, r = nums.length - 1;
int idx = 0;

while (idx != nums.length) {
res[idx++] = nums[l++];
if (l <= r) {
res[idx++] = nums[r--];
}
}

return res;
}
}
```

```cpp
class Solution {
public:
vector<int> rearrangeArray(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<int> res;
int l = 0, r = nums.size() - 1;

while (res.size() != nums.size()) {
res.push_back(nums[l++]);
if (l <= r) {
res.push_back(nums[r--]);
}
}

return res;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} nums
* @return {number[]}
*/
rearrangeArray(nums) {
nums.sort((a, b) => a - b);
const res = [];
let l = 0, r = nums.length - 1;

while (res.length !== nums.length) {
res.push(nums[l++]);
if (l <= r) {
res.push(nums[r--]);
}
}

return res;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n\log n)$
* Space complexity: $O(n)$

---

## 2. Greedy (Space Optimized)

::tabs-start

```python
class Solution:
def rearrangeArray(self, nums: List[int]) -> List[int]:
nums.sort()
for i in range(1, len(nums), 2):
nums[i], nums[i - 1] = nums[i - 1], nums[i]
return nums
```

```java
public class Solution {
public int[] rearrangeArray(int[] nums) {
Arrays.sort(nums);
for (int i = 1; i < nums.length; i += 2) {
int temp = nums[i];
nums[i] = nums[i - 1];
nums[i - 1] = temp;
}
return nums;
}
}
```

```cpp
class Solution {
public:
vector<int> rearrangeArray(vector<int>& nums) {
sort(nums.begin(), nums.end());
for (int i = 1; i < nums.size(); i += 2) {
swap(nums[i], nums[i - 1]);
}
return nums;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} nums
* @return {number[]}
*/
rearrangeArray(nums) {
nums.sort((a, b) => a - b);
for (let i = 1; i < nums.length; i += 2) {
[nums[i], nums[i - 1]] = [nums[i - 1], nums[i]];
}
return nums;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n \log n)$
* Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

---

## 3. Greedy (Optimal) - I

::tabs-start

```python
class Solution:
def rearrangeArray(self, nums: List[int]) -> List[int]:
n = len(nums)

for i in range(1, n - 1):
if 2 * nums[i] == (nums[i - 1] + nums[i + 1]):
nums[i], nums[i + 1] = nums[i + 1], nums[i]

for i in range(n - 2, 0, -1):
if 2 * nums[i] == (nums[i - 1] + nums[i + 1]):
nums[i], nums[i - 1] = nums[i - 1], nums[i]

return nums
```

```java
public class Solution {
public int[] rearrangeArray(int[] nums) {
int n = nums.length;

for (int i = 1; i < n - 1; i++) {
if (2 * nums[i] == (nums[i - 1] + nums[i + 1])) {
int temp = nums[i];
nums[i] = nums[i + 1];
nums[i + 1] = temp;
}
}

for (int i = n - 2; i > 0; i--) {
if (2 * nums[i] == (nums[i - 1] + nums[i + 1])) {
int temp = nums[i];
nums[i] = nums[i - 1];
nums[i - 1] = temp;
}
}

return nums;
}
}
```

```cpp
class Solution {
public:
vector<int> rearrangeArray(vector<int>& nums) {
int n = nums.size();

for (int i = 1; i < n - 1; i++) {
if (2 * nums[i] == (nums[i - 1] + nums[i + 1])) {
swap(nums[i], nums[i + 1]);
}
}

for (int i = n - 2; i > 0; i--) {
if (2 * nums[i] == (nums[i - 1] + nums[i + 1])) {
swap(nums[i], nums[i - 1]);
}
}

return nums;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} nums
* @return {number[]}
*/
rearrangeArray(nums) {
const n = nums.length;

for (let i = 1; i < n - 1; i++) {
if (2 * nums[i] === nums[i - 1] + nums[i + 1]) {
[nums[i], nums[i + 1]] = [nums[i + 1], nums[i]];
}
}

for (let i = n - 2; i > 0; i--) {
if (2 * nums[i] === nums[i - 1] + nums[i + 1]) {
[nums[i], nums[i - 1]] = [nums[i - 1], nums[i]];
}
}

return nums;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n)$
* Space complexity: $O(1)$ extra space.

---

## 4. Greedy Optimal - II

::tabs-start

```python
class Solution:
def rearrangeArray(self, nums: List[int]) -> List[int]:
increase = nums[0] < nums[1]
for i in range(1, len(nums) - 1):
if ((increase and nums[i] < nums[i + 1]) or
(not increase and nums[i] > nums[i + 1])
):
nums[i], nums[i + 1] = nums[i + 1], nums[i]
increase = not increase
return nums
```

```java
public class Solution {
public int[] rearrangeArray(int[] nums) {
boolean increase = nums[0] < nums[1];
for (int i = 1; i < nums.length - 1; i++) {
if ((increase && nums[i] < nums[i + 1]) ||
(!increase && nums[i] > nums[i + 1])) {
int temp = nums[i];
nums[i] = nums[i + 1];
nums[i + 1] = temp;
}
increase = !increase;
}
return nums;
}
}
```

```cpp
class Solution {
public:
vector<int> rearrangeArray(vector<int>& nums) {
bool increase = nums[0] < nums[1];
for (int i = 1; i < nums.size() - 1; i++) {
if ((increase && nums[i] < nums[i + 1]) ||
(!increase && nums[i] > nums[i + 1])) {
swap(nums[i], nums[i + 1]);
}
increase = !increase;
}
return nums;
}
};
```

```javascript
class Solution {
/**
* @param {number[]} nums
* @return {number[]}
*/
rearrangeArray(nums) {
let increase = nums[0] < nums[1];
for (let i = 1; i < nums.length - 1; i++) {
if ((increase && nums[i] < nums[i + 1]) ||
(!increase && nums[i] > nums[i + 1])) {
[nums[i], nums[i + 1]] = [nums[i + 1], nums[i]];
}
increase = !increase;
}
return nums;
}
}
```

::tabs-end

### Time & Space Complexity

* Time complexity: $O(n)$
* Space complexity: $O(1)$ extra space.
Loading