709904. Even Swap
Here is the Markdown blog post for the "Even Swap" problem:
Even Swap
Summary
The Even Swap problem requires you to sort an array of integers in such a way that all even numbers are swapped with their corresponding even numbers on the right side of the array. The key concept involved is array manipulation and sorting.
Detailed Explanation
To solve this problem, we can iterate through the array and keep track of the indices where even numbers are found. We then swap these even numbers with their corresponding even numbers to the right in a separate array. Once all swaps have been made, we concatenate the original array (excluding the even numbers) with the swapped even numbers to get the final sorted array.
Here is the step-by-step breakdown of the solution:
- Initialize two arrays:
evenandodd.evenwill store the indices of even numbers in the original array, whileoddwill store the odd numbers. - Iterate through the original array:
- For each number, check if it's even or odd.
- If it's even, add its index to the
evenarray. - If it's odd, add the number to the
oddarray.
- Swap the elements at indices in the
evenandoddarrays:- For each index in
even, swap the element at that index with the element at the corresponding index from the end of theoddarray (since we're working backwards).
- For each index in
- Concatenate the original array (excluding the even numbers) with the swapped even numbers to get the final sorted array.
Time complexity: O(n), where n is the length of the array. Space complexity: O(1), since we only use a constant amount of space to store the indices and arrays.
Optimized Solutions
Java
public int[] evenSwap(int[] arr) {
int evenIndex = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] % 2 == 0) {
evenIndex++;
} else {
// store odd numbers in a separate array
arr[evenIndex++] = arr[i];
}
}
int oddIndex = 0;
for (int i = arr.length - 1; i >= 0; i--) {
if (arr[i] % 2 != 0) {
arr[i] = arr[--oddIndex];
}
}
return arr;
}