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.
Key Insights
- The problem requires separating even and odd integers in an array.
- The order of even and odd integers relative to each other does not matter.
- A two-pointer approach can be effective in solving this problem efficiently.
Space and Time Complexity
Time Complexity: O(n) - We need to iterate through the array once. Space Complexity: O(1) - We can perform the operations in place without using extra space.
Solution
To solve the problem, we can use a two-pointer approach. We maintain two pointers, one starting from the beginning of the array and the other from the end. The first pointer will move forward until it finds an odd number, and the second pointer will move backward until it finds an even number. When both pointers meet, we swap the elements at these pointers, continuing the process until the pointers cross. This effectively groups all even numbers at the front and odd numbers at the back.