# Sort an array of 0s, 1s and 2s

Problem Statement: Given an array consisting of only 0s, 1s and 2s. Write a program to in-place sort the array without using inbuilt sort functions. ( Expected: Single pass-O(N) and constant space)

Example 1:

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

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

Input: nums = 
Input: nums = ```

### Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1: Sorting ( even if it is not the expected solution here but it can be considered as one of the approaches). Since the array contains only 3 integers, 0, 1, and 2. Simply sorting the array would arrange the elements in increasing order.

Time Complexity: O(N*logN)

Space Complexity: O(1)

Solution 2: Keeping count of values

Intuition: Since in this case there are only 3 distinct values in the array so it’s easy to maintain the count of all, Like the count of 0, 1, and 2. This can be followed by overwriting the array based on the frequency(count) of the values.

Approach:

1. Take 3 variables to maintain the count of 0, 1 and 2.
2. Travel the array once and increment the corresponding counting variables

( let’s consider count_0 = a, count_1 = b, count_2 = c )

1. In 2nd traversal of array, we will now over write the first ‘a’ indices / positions in array with ’0’, the next ‘b’ with ‘1’ and the remaining ‘c’ with ‘2’.

Time Complexity: O(N) + O(N)

Space Complexity: O(1)

Solution 3: 3-Pointer approach

This problem is a variation of the popular Dutch National flag algorithm

Intuition: In this approach, we will be using 3 pointers named low, mid, and high. We will be using these 3 pointers to move around the values. The primary goal here is to move 0s to the left and 2s to the right of the array and at the same time all the 1s shall be in the middle region of the array and hence the array will be sorted.

Approach:

1. Initialize the 3 pointers such that low and mid will point to 0th index and high pointer will point to last index

int low = arr

int mid = arr

int high = arr[n – 1]

1. Now there will 3 different operations / steps based on the value of arr[mid] and will be repeated until mid <= high.

1. arr[mid] == 0:
swap(arr[low], arr[mid])
low++, mid++

1. arr[mid] == 1:
mid++

1. arr[mid] == 2:
swap(arr[mid], arr[high])
high–;

The array formed after these steps will be a sorted array.

## C++ Code

``````class Solution {
public:
void sortColors(vector<int>& nums) {
int lo = 0;
int hi = nums.size() - 1;
int mid = 0;

while (mid <= hi) {
switch (nums[mid]) {

// If the element is 0
case 0:
swap(nums[lo++], nums[mid++]);
break;

// If the element is 1 .
case 1:
mid++;
break;

// If the element is 2
case 2:
swap(nums[mid], nums[hi--]);
break;
}
}

}
};``````

## Java Code

``````class Solution {
public void sortColors(int[] nums) {
int lo = 0;
int hi = nums.length - 1;
int mid = 0;
int temp;
while (mid <= hi) {
switch (nums[mid]) {
case 0: {
temp = nums[lo];
nums[lo] = nums[mid];
nums[mid] = temp;
lo++;
mid++;
break;
}
case 1:
mid++;
break;
case 2: {
temp = nums[mid];
nums[mid] = nums[hi];
nums[hi] = temp;
hi--;
break;
}
}
}
}
}``````

Time Complexity: O(N)

Space Complexity: O(1)

Special thanks to Aditya Shahare for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article.