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Problem

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.

Examples

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]

Solution

Method 1 - Two Pointer Partitioning

Intuition

We want to move all even numbers to the front and all odd numbers to the back. By using two pointers, we can swap misplaced elements in-place, efficiently partitioning the array.

Approach

Initialize two pointers: one at the start (l), one at the end (r).
Move l forward if it points to an even number.
If l points to an odd number, swap it with the element at r and move r backward.
Continue until l meets r.
Return the modified array.

Complexity

⏰ Time complexity: O(n) — Each element is checked at most once.
🧺 Space complexity: O(1) — In-place swaps, no extra space.

Code

[{"language":"Java","code":"class Solution {\n\tpublic int[] sortArrayByParity(int[] nums) {\n\t\tint l = 0, r = nums.length - 1;\n\t\twhile (l < r) {\n\t\t\tif (nums[l] % 2 == 0) {\n\t\t\t\tl++;\n\t\t\t} else {\n\t\t\t\tint temp = nums[l];\n\t\t\t\tnums[l] = nums[r];\n\t\t\t\tnums[r] = temp;\n\t\t\t\tr--;\n\t\t\t}\n\t\t}\n\t\treturn nums;\n\t}\n}","lang":"java"},{"language":"Python","code":"class Solution:\n\tdef sortArrayByParity(self, nums):\n\t\tl, r = 0, len(nums) - 1\n\t\twhile l < r:\n\t\t\tif nums[l] % 2 == 0:\n\t\t\t\tl += 1\n\t\t\telse:\n\t\t\t\tnums[l], nums[r] = nums[r], nums[l]\n\t\t\t\tr -= 1\n\t\treturn nums","lang":"python"},{"language":"C++","code":"class Solution {\npublic:\n\tvector<int> sortArrayByParity(vector<int>& nums) {\n\t\tint l = 0, r = nums.size() - 1;\n\t\twhile (l < r) {\n\t\t\tif (nums[l] % 2 == 0) {\n\t\t\t\tl++;\n\t\t\t} else {\n\t\t\t\tswap(nums[l], nums[r]);\n\t\t\t\tr--;\n\t\t\t}\n\t\t}\n\t\treturn nums;\n\t}\n};","lang":"cpp"},{"language":"Go","code":"func sortArrayByParity(nums []int) []int {\n\tl, r := 0, len(nums)-1\n\tfor l < r {\n\t\tif nums[l]%2 == 0 {\n\t\t\tl++\n\t\t} else {\n\t\t\tnums[l], nums[r] = nums[r], nums[l]\n\t\t\tr--\n\t\t}\n\t}\n\treturn nums\n}","lang":"go"},{"language":"Kotlin","code":"class Solution {\n\tfun sortArrayByParity(nums: IntArray): IntArray {\n\t\tvar l = 0\n\t\tvar r = nums.size - 1\n\t\twhile (l < r) {\n\t\t\tif (nums[l] % 2 == 0) {\n\t\t\t\tl++\n\t\t\t} else {\n\t\t\t\tval tmp = nums[l]\n\t\t\t\tnums[l] = nums[r]\n\t\t\t\tnums[r] = tmp\n\t\t\t\tr--\n\t\t\t}\n\t\t}\n\t\treturn nums\n\t}\n}","lang":"kotlin"},{"language":

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