# 75. Sort Colors

## Description

Given an array nums with n objects colored red, white, or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white, and blue.

We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.

You must solve this problem without using the library's sort function.

Example 1:

Input: nums = [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]


Example 2:

Input: nums = [2,0,1]
Output: [0,1,2]


Constraints:

• n == nums.length
• 1 <= n <= 300
• nums[i] is either 0, 1, or 2.

Follow up: Could you come up with a one-pass algorithm using only constant extra space?

## Solutions

Solution 1: Three Pointers

We define three pointers $i$, $j$, and $k$. Pointer $i$ is used to point to the rightmost boundary of the elements with a value of $0$ in the array, and pointer $j$ is used to point to the leftmost boundary of the elements with a value of $2$ in the array. Initially, $i=-1$, $j=n$. Pointer $k$ is used to point to the current element being traversed, initially $k=0$.

When $k < j$, we perform the following operations:

• If $nums[k] = 0$, then swap it with $nums[i+1]$, then increment both $i$ and $k$ by $1$;
• If $nums[k] = 2$, then swap it with $nums[j-1]$, then decrement $j$ by $1$;
• If $nums[k] = 1$, then increment $k$ by $1$.

After the traversal, the elements in the array are divided into three parts: $[0,i]$, $[i+1,j-1]$ and $[j,n-1]$.

The time complexity is $O(n)$, where $n$ is the length of the array. Only one traversal of the array is needed. The space complexity is $O(1)$.

• class Solution {
public void sortColors(int[] nums) {
int i = -1, j = nums.length, k = 0;
while (k < j) {
if (nums[k] == 0) {
swap(nums, ++i, k++);
} else if (nums[k] == 2) {
swap(nums, --j, k);
} else {
++k;
}
}
}

private void swap(int[] nums, int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}

• class Solution {
public:
void sortColors(vector<int>& nums) {
int i = -1, j = nums.size(), k = 0;
while (k < j) {
if (nums[k] == 0) {
swap(nums[++i], nums[k++]);
} else if (nums[k] == 2) {
swap(nums[--j], nums[k]);
} else {
++k;
}
}
}
};

• class Solution:
def sortColors(self, nums: List[int]) -> None:
i, j, k = -1, len(nums), 0
while k < j:
if nums[k] == 0:
i += 1
nums[i], nums[k] = nums[k], nums[i]
k += 1
elif nums[k] == 2:
j -= 1
nums[j], nums[k] = nums[k], nums[j]
else:
k += 1


• func sortColors(nums []int) {
i, j, k := -1, len(nums), 0
for k < j {
if nums[k] == 0 {
i++
nums[i], nums[k] = nums[k], nums[i]
k++
} else if nums[k] == 2 {
j--
nums[j], nums[k] = nums[k], nums[j]
} else {
k++
}
}
}

• /**
Do not return anything, modify nums in-place instead.
*/
function sortColors(nums: number[]): void {
let i = -1;
let j = nums.length;
let k = 0;
while (k < j) {
if (nums[k] === 0) {
++i;
[nums[i], nums[k]] = [nums[k], nums[i]];
++k;
} else if (nums[k] === 2) {
--j;
[nums[j], nums[k]] = [nums[k], nums[j]];
} else {
++k;
}
}
}


• public class Solution {
public void SortColors(int[] nums) {
int i = -1, j = nums.Length, k = 0;
while (k < j) {
if (nums[k] == 0) {
swap(nums, ++i, k++);
} else if (nums[k] == 2) {
swap(nums, --j, k);
} else {
++k;
}
}
}

private void swap(int[] nums, int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}

• impl Solution {
pub fn sort_colors(nums: &mut Vec<i32>) {
let mut i = -1;
let mut j = nums.len();
let mut k = 0;
while k < j {
if nums[k] == 0 {
i += 1;
nums.swap(i as usize, k as usize);
k += 1;
} else if nums[k] == 2 {
j -= 1;
nums.swap(j, k);
} else {
k += 1;
}
}
}
}