Merge pull request #3833 from katarianikita2003/main

Create 0905-Sort-Array-By-Parity.md

Author: ajay-dhangar

dsa-solutions/lc-solutions/0900-0999/0905-Sort-Array-By-Parity.md

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+---
+id: Sort-Array-By-Parity
+title: Sort Array By Parity
+sidebar_label: Sort Array By Parity
+tags: 
+    - Arrays
+    - Sorting
+---
+
+## Problem Description
+
+| Problem Statement                                       | Solution Link                                                              | LeetCode Profile                                        |
+| :------------------------------------------------------ | :------------------------------------------------------------------------- | :------------------------------------------------------ |
+| [Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/description/) | [Sort Array By Parity Solution on LeetCode](https://leetcode.com/problems/sort-array-by-parity/solutions/) | [Nikita Saini](https://leetcode.com/u/Saini_Nikita/) |
+
+## Problem Description
+
+Given an integer array `nums`, move all the even integers at the beginning of the array followed by all the odd integers.
+
+Return any array that satisfies this condition.
+
+### Example 1:
+
+**Input:** `nums = [3, 1, 2, 4]`  
+**Output:** `[2, 4, 3, 1]`  
+**Explanation:** The outputs `[4, 2, 3, 1]`, `[2, 4, 1, 3]`, and `[4, 2, 1, 3]` would also be accepted.
+
+### Example 2:
+
+**Input:** `nums = [0]`  
+**Output:** `[0]`
+
+## Constraints
+
+- `1 <= nums.length <= 5000`
+- `0 <= nums[i] <= 5000`
+
+## Approach
+
+1. **Identify even and odd integers**: Iterate through the array and separate the even and odd integers.
+2. **Rearrange the array**: Place all even integers at the beginning of the array followed by the odd integers.
+
+## Solution
+
+### Python
+
+```python
+def sortArrayByParity(nums):
+    even = [x for x in nums if x % 2 == 0]
+    odd = [x for x in nums if x % 2 != 0]
+    return even + odd
+
+# Example usage
+nums = [3, 1, 2, 4]
+print(sortArrayByParity(nums))  # Output: [2, 4, 3, 1]
+```
+
+### Java
+
+```java
+import java.util.*;
+
+public class EvenOddArray {
+    public static int[] sortArrayByParity(int[] nums) {
+        List<Integer> even = new ArrayList<>();
+        List<Integer> odd = new ArrayList<>();
+        
+        for (int num : nums) {
+            if (num % 2 == 0) {
+                even.add(num);
+            } else {
+                odd.add(num);
+            }
+        }
+        
+        even.addAll(odd);
+        return even.stream().mapToInt(i -> i).toArray();
+    }
+    
+    public static void main(String[] args) {
+        int[] nums = {3, 1, 2, 4};
+        System.out.println(Arrays.toString(sortArrayByParity(nums)));  // Output: [2, 4, 3, 1]
+    }
+}
+```
+
+### C++
+
+```cpp
+#include <vector>
+#include <iostream>
+
+std::vector<int> sortArrayByParity(std::vector<int>& nums) {
+    std::vector<int> even, odd;
+    for (int num : nums) {
+        if (num % 2 == 0) {
+            even.push_back(num);
+        } else {
+            odd.push_back(num);
+        }
+    }
+    even.insert(even.end(), odd.begin(), odd.end());
+    return even;
+}
+
+int main() {
+    std::vector<int> nums = {3, 1, 2, 4};
+    std::vector<int> result = sortArrayByParity(nums);
+    for (int num : result) {
+        std::cout << num << " ";
+    }
+    // Output: 2 4 3 1
+}
+```
+
+### C
+
+```c
+#include <stdio.h>
+#include <stdlib.h>
+
+void sortArrayByParity(int* nums, int numsSize, int* returnSize) {
+    int* result = (int*)malloc(numsSize * sizeof(int));
+    int evenIndex = 0, oddIndex = numsSize - 1;
+    
+    for (int i = 0; i < numsSize; ++i) {
+        if (nums[i] % 2 == 0) {
+            result[evenIndex++] = nums[i];
+        } else {
+            result[oddIndex--] = nums[i];
+        }
+    }
+    *returnSize = numsSize;
+    for (int i = 0; i < numsSize; ++i) {
+        nums[i] = result[i];
+    }
+    free(result);
+}
+
+int main() {
+    int nums[] = {3, 1, 2, 4};
+    int numsSize = sizeof(nums) / sizeof(nums[0]);
+    int returnSize;
+    sortArrayByParity(nums, numsSize, &returnSize);
+    
+    for (int i = 0; i < numsSize; ++i) {
+        printf("%d ", nums[i]);
+    }
+    // Output: 2 4 3 1
+    return 0;
+}
+```
+
+### JavaScript
+
+```javascript
+function sortArrayByParity(nums) {
+    let even = [];
+    let odd = [];
+    
+    for (let num of nums) {
+        if (num % 2 === 0) {
+            even.push(num);
+        } else {
+            odd.push(num);
+        }
+    }
+    
+    return [...even, ...odd];
+}
+
+// Example usage
+let nums = [3, 1, 2, 4];
+console.log(sortArrayByParity(nums));  // Output: [2, 4, 3, 1]
+```
+
+## Step-by-Step Algorithm
+
+1. Initialize two empty lists/arrays: one for even integers and one for odd integers.
+2. Iterate through the given array:
+   - If the current integer is even, add it to the even list/array.
+   - If the current integer is odd, add it to the odd list/array.
+3. Concatenate the even list/array with the odd list/array.
+4. Return the concatenated list/array.
+
+## Conclusion
+
+This problem can be solved efficiently by iterating through the array once and separating the integers into even and odd lists/arrays. The time complexity is O(n), where n is the length of the array, making this approach optimal for the given constraints.