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

Given an array A[] consisting 0s, 1s and 2s. The task is to write a function that sorts the given array. The functions should put all 0s first, then all 1s and all 2s in last.

Examples:

```Input: {0, 1, 2, 0, 1, 2}
Output: {0, 0, 1, 1, 2, 2}

Input: {0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1}
Output: {0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2}
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A simple solution is discussed in this(Sort an array of 0s, 1s and 2s (Simple Counting)) post.

Method 1

• Approach:The problem is similar to our old post Segregate 0s and 1s in an array, and both of these problems are variation of famous Dutch national flag problem.

The problem was posed with three colours, here `0′, `1′ and `2′. The array is divided into four sections:

1. a[1..Lo-1] zeroes (red)
2. a[Lo..Mid-1] ones (white)
3. a[Mid..Hi] unknown
4. a[Hi+1..N] twos (blue)

Dutch National Flag Algorithm OR 3-way Partitioning:
At first, the full array is unknown. There are three indices – low, mid and high. Initially, the value of low = mid = 1 and high = N.

• If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range.
• Similarly, if the element is 1 then keep it as it is but shrink the unknown range.
• If the element is 2 then swap it with an element in high range.
• Algorithm:

1. Keep three indices low = 1, mid = 1 and high = N and there are four ranges, 1 to low (the range containing 0), low to mid (the range containing 1), mid to high (the range containing unknown elements) and high to N (the range containing 2).
2. Traverse the array from start to end and mid is less than high. (Loop counter is i)
3. If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1
4. If the element is 1 then update mid = mid + 1
5. If the element is 2 then swap the element with the element at index high and update high = high – 1 and update i = i – 1. As the swapped element is not processed
6. Print the output array.
• Dry Run:
Part way through the process, some red, white and blue elements are known and are in the “right” place. The section of unknown elements, a[Mid..Hi], is shrunk by examining a[Mid]: Examine a[Mid]. There are three possibilities:
a[Mid] is (0) red, (1) white or (2) blue.
Case (0) a[Mid] is red, swap a[Lo] and a[Mid]; Lo++; Mid++ Case (1) a[Mid] is white, Mid++ Case (2) a[Mid] is blue, swap a[Mid] and a[Hi]; Hi– Continue until Mid>Hi.

• Implementation:

## C++

 `// C++ program to sort an array ` `// with 0, 1 and 2 in a single pass ` `#include ` `using` `namespace` `std; ` ` `  `// Function to sort the input array, ` `// the array is assumed ` `// to have values in {0, 1, 2} ` `void` `sort012(``int` `a[], ``int` `arr_size) ` `{ ` `    ``int` `lo = 0; ` `    ``int` `hi = arr_size - 1; ` `    ``int` `mid = 0; ` ` `  `    ``// Iterate till all the elements ` `    ``// are sorted ` `    ``while` `(mid <= hi) { ` `        ``switch` `(a[mid]) { ` ` `  `        ``// If the element is 0 ` `        ``case` `0: ` `            ``swap(a[lo++], a[mid++]); ` `            ``break``; ` ` `  `        ``// If the element is 1 . ` `        ``case` `1: ` `            ``mid++; ` `            ``break``; ` ` `  `        ``// If the element is 2 ` `        ``case` `2: ` `            ``swap(a[mid], a[hi--]); ` `            ``break``; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to print array arr[] ` `void` `printArray(``int` `arr[], ``int` `arr_size) ` `{ ` `    ``// Iterate and print every element ` `    ``for` `(``int` `i = 0; i < arr_size; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``sort012(arr, n); ` ` `  `    ``cout << ``"array after segregation "``; ` ` `  `    ``printArray(arr, n); ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by Shivi_Aggarwal `

## Javascript

 `                     `

## C

 `// C program to sort an array with 0, 1 and 2 ` `// in a single pass ` `#include ` ` `  `/* Function to swap *a and *b */` `void` `swap(``int``* a, ``int``* b); ` ` `  `// Sort the input array, the array is assumed to ` `// have values in {0, 1, 2} ` `void` `sort012(``int` `a[], ``int` `arr_size) ` `{ ` `    ``int` `lo = 0; ` `    ``int` `hi = arr_size - 1; ` `    ``int` `mid = 0; ` ` `  `    ``while` `(mid <= hi) { ` `        ``switch` `(a[mid]) { ` `        ``case` `0: ` `            ``swap(&a[lo++], &a[mid++]); ` `            ``break``; ` `        ``case` `1: ` `            ``mid++; ` `            ``break``; ` `        ``case` `2: ` `            ``swap(&a[mid], &a[hi--]); ` `            ``break``; ` `        ``} ` `    ``} ` `} ` ` `  `/* UTILITY FUNCTIONS */` `void` `swap(``int``* a, ``int``* b) ` `{ ` `    ``int` `temp = *a; ` `    ``*a = *b; ` `    ``*b = temp; ` `} ` ` `  `/* Utility function to print array arr[] */` `void` `printArray(``int` `arr[], ``int` `arr_size) ` `{ ` `    ``int` `i; ` `    ``for` `(i = 0; i < arr_size; i++) ` `        ``printf``(``"%d "``, arr[i]); ` `    ``printf``(``"n"``); ` `} ` ` `  `/* driver program to test */` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 }; ` `    ``int` `arr_size = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `i; ` ` `  `    ``sort012(arr, arr_size); ` ` `  `    ``printf``(``"array after segregation "``); ` `    ``printArray(arr, arr_size); ` ` `  `    ``getchar``(); ` `    ``return` `0; ` `} `

## Java

 `// Java program to sort an array of 0, 1 and 2 ` `import` `java.io.*; ` ` `  `class` `countzot { ` ` `  `    ``// Sort the input array, the array is assumed to ` `    ``// have values in {0, 1, 2} ` `    ``static` `void` `sort012(``int` `a[], ``int` `arr_size) ` `    ``{ ` `        ``int` `lo = ``0``; ` `        ``int` `hi = arr_size - ``1``; ` `        ``int` `mid = ``0``, temp = ``0``; ` `        ``while` `(mid <= hi) { ` `            ``switch` `(a[mid]) { ` `            ``case` `0``: { ` `                ``temp = a[lo]; ` `                ``a[lo] = a[mid]; ` `                ``a[mid] = temp; ` `                ``lo++; ` `                ``mid++; ` `                ``break``; ` `            ``} ` `            ``case` `1``: ` `                ``mid++; ` `                ``break``; ` `            ``case` `2``: { ` `                ``temp = a[mid]; ` `                ``a[mid] = a[hi]; ` `                ``a[hi] = temp; ` `                ``hi--; ` `                ``break``; ` `            ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``/* Utility function to print array arr[] */` `    ``static` `void` `printArray(``int` `arr[], ``int` `arr_size) ` `    ``{ ` `        ``int` `i; ` `        ``for` `(i = ``0``; i < arr_size; i++) ` `            ``System.out.print(arr[i] + ``" "``); ` `        ``System.out.println(``""``); ` `    ``} ` ` `  `    ``/*Driver function to check for above functions*/` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``0``, ``1``, ``1``, ``0``, ``1``, ``2``, ``1``, ``2``, ``0``, ``0``, ``0``, ``1` `}; ` `        ``int` `arr_size = arr.length; ` `        ``sort012(arr, arr_size); ` `        ``System.out.println(``"Array after seggregation "``); ` `        ``printArray(arr, arr_size); ` `    ``} ` `} ` `/*This code is contributed by Devesh Agrawal*/`

## Python

 `# Python program to sort an array with  ` `# 0, 1 and 2 in a single pass ` ` `  `# Function to sort array ` `def` `sort012( a, arr_size): ` `    ``lo ``=` `0` `    ``hi ``=` `arr_size ``-` `1` `    ``mid ``=` `0` `    ``while` `mid <``=` `hi: ` `        ``if` `a[mid] ``=``=` `0``: ` `            ``a[lo], a[mid] ``=` `a[mid], a[lo] ` `            ``lo ``=` `lo ``+` `1` `            ``mid ``=` `mid ``+` `1` `        ``elif` `a[mid] ``=``=` `1``: ` `            ``mid ``=` `mid ``+` `1` `        ``else``: ` `            ``a[mid], a[hi] ``=` `a[hi], a[mid]  ` `            ``hi ``=` `hi ``-` `1` `return` `a ` `     `  `# Function to print array ` `def` `printArray( a): ` `    ``for` `k ``in` `a: ` `        ``print` `k, ` `    ``print` `     `  ` `  `# Driver Program ` `arr ``=` `[``0``, ``1``, ``1``, ``0``, ``1``, ``2``, ``1``, ``2``, ``0``, ``0``, ``0``, ``1``] ` `arr_size ``=` `len``(arr) ` `arr ``=` `sort012( arr, arr_size) ` `print` `"Array aftter segregation :"``, ` `printArray(arr) ` ` `  `# Contributed by Harshit Agrawal `

## C#

 `// C# program to sort an ` `// array of 0, 1 and 2 ` `using` `System; ` ` `  `class` `GFG { ` `    ``// Sort the input array, the array is assumed to ` `    ``// have values in {0, 1, 2} ` `    ``static` `void` `sort012(``int``[] a, ``int` `arr_size) ` `    ``{ ` `        ``int` `lo = 0; ` `        ``int` `hi = arr_size - 1; ` `        ``int` `mid = 0, temp = 0; ` ` `  `        ``while` `(mid <= hi) { ` `            ``switch` `(a[mid]) { ` `            ``case` `0: { ` `                ``temp = a[lo]; ` `                ``a[lo] = a[mid]; ` `                ``a[mid] = temp; ` `                ``lo++; ` `                ``mid++; ` `                ``break``; ` `            ``} ` `            ``case` `1: ` `                ``mid++; ` `                ``break``; ` `            ``case` `2: { ` `                ``temp = a[mid]; ` `                ``a[mid] = a[hi]; ` `                ``a[hi] = temp; ` `                ``hi--; ` `                ``break``; ` `            ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``/* Utility function to print array arr[] */` `    ``static` `void` `printArray(``int``[] arr, ``int` `arr_size) ` `    ``{ ` `        ``int` `i; ` ` `  `        ``for` `(i = 0; i < arr_size; i++) ` `            ``Console.Write(arr[i] + ``" "``); ` `        ``Console.WriteLine(``""``); ` `    ``} ` ` `  `    ``/*Driver function to check for above functions*/` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 }; ` `        ``int` `arr_size = arr.Length; ` `        ``sort012(arr, arr_size); ` ` `  `        ``Console.Write(``"Array after seggregation "``); ` ` `  `        ``printArray(arr, arr_size); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

Output:

```array after segregation
0 0 0 0 0 1 1 1 1 1 2 2 ```
• Complexity Analysis:

• Time Complexity: O(n).
Only one traversal of the array is needed.
• Space Complexity: O(1).
No extra space is required.
• The above code performs unnecessary swaps for some inputs which are not really required. It can be modified to reduce some swaps.

Thanks to Ankur Roy for suggesting this optimization.

Method 2

• Approach: Count the number of 0s, 1s and 2s in the given array. Then store all the 0s in the beginning followed by all the 1s then all the 2s.
• Algorithm:
1. Keep three counter c0 to count 0s, c1 to count 1s and c2 to count 2s
2. Traverse through the array and increase the count of c0 is the element is 0,increase the count of c1 is the element is 1 and increase the count of c2 is the element is 2
3. Now again traverse the array and replace first c0 elements with 0, next c1 elements with 1 and next c2 elements with 2.
• Implementation:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to print the contents of an array ` `void` `printArr(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Function to sort the array of 0s, 1s and 2s ` `void` `sortArr(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `i, cnt0 = 0, cnt1 = 0, cnt2 = 0; ` ` `  `    ``// Count the number of 0s, 1s and 2s in the array ` `    ``for` `(i = 0; i < n; i++) { ` `        ``switch` `(arr[i]) { ` `        ``case` `0: ` `            ``cnt0++; ` `            ``break``; ` `        ``case` `1: ` `            ``cnt1++; ` `            ``break``; ` `        ``case` `2: ` `            ``cnt2++; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Update the array ` `    ``i = 0; ` ` `  `    ``// Store all the 0s in the beginning ` `    ``while` `(cnt0 > 0) { ` `        ``arr[i++] = 0; ` `        ``cnt0--; ` `    ``} ` ` `  `    ``// Then all the 1s ` `    ``while` `(cnt1 > 0) { ` `        ``arr[i++] = 1; ` `        ``cnt1--; ` `    ``} ` ` `  `    ``// Finally all the 2s ` `    ``while` `(cnt2 > 0) { ` `        ``arr[i++] = 2; ` `        ``cnt2--; ` `    ``} ` ` `  `    ``// Print the sorted array ` `    ``printArr(arr, n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``sortArr(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG { ` `    ``// Utility function to print the contents of an array ` `    ``static` `void` `printArr(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``System.out.print(arr[i] + ``" "``); ` `    ``} ` `     `  `    ``// Function to sort the array of 0s, 1s and 2s ` `    ``static` `void` `sortArr(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `i, cnt0 = ``0``, cnt1 = ``0``, cnt2 = ``0``; ` `     `  `        ``// Count the number of 0s, 1s and 2s in the array ` `        ``for` `(i = ``0``; i < n; i++) { ` `            ``switch` `(arr[i]) { ` `            ``case` `0``: ` `                ``cnt0++; ` `                ``break``; ` `            ``case` `1``: ` `                ``cnt1++; ` `                ``break``; ` `            ``case` `2``: ` `                ``cnt2++; ` `                ``break``; ` `            ``} ` `        ``} ` `     `  `        ``// Update the array ` `        ``i = ``0``; ` `     `  `        ``// Store all the 0s in the beginning ` `        ``while` `(cnt0 > ``0``) { ` `            ``arr[i++] = ``0``; ` `            ``cnt0--; ` `        ``} ` `     `  `        ``// Then all the 1s ` `        ``while` `(cnt1 > ``0``) { ` `            ``arr[i++] = ``1``; ` `            ``cnt1--; ` `        ``} ` `     `  `        ``// Finally all the 2s ` `        ``while` `(cnt2 > ``0``) { ` `            ``arr[i++] = ``2``; ` `            ``cnt2--; ` `        ``} ` `     `  `        ``// Print the sorted array ` `        ``printArr(arr, n); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``0``, ``1``, ``1``, ``0``, ``1``, ``2``, ``1``, ``2``, ``0``, ``0``, ``0``, ``1` `}; ` `        ``int` `n = arr.length; ` `        ``sortArr(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by shubhamsingh10 `

## Python

 `# Python implementation of the approach ` ` `  `# Utility function to prthe contents of an array ` `def` `printArr(arr, n): ` `    ``for` `i ``in` `range``(n): ` `        ``print``(arr[i],end``=``" "``) ` ` `  ` `  `# Function to sort the array of 0s, 1s and 2s ` `def` `sortArr(arr, n): ` `    ``cnt0 ``=` `0` `    ``cnt1 ``=` `0` `    ``cnt2 ``=` `0` `     `  `    ``# Count the number of 0s, 1s and 2s in the array ` `    ``for` `i ``in` `range``(n): ` `        ``if` `arr[i] ``=``=` `0``: ` `            ``cnt0``+``=``1` `         `  `        ``elif` `arr[i] ``=``=` `1``: ` `            ``cnt1``+``=``1` `             `  `        ``elif` `arr[i] ``=``=` `2``: ` `            ``cnt2``+``=``1` `     `  `    ``# Update the array ` `    ``i ``=` `0` `     `  `    ``# Store all the 0s in the beginning ` `    ``while` `(cnt0 > ``0``): ` `        ``arr[i] ``=` `0` `        ``i``+``=``1` `        ``cnt0``-``=``1` `     `  `    ``# Then all the 1s ` `    ``while` `(cnt1 > ``0``): ` `        ``arr[i] ``=` `1` `        ``i``+``=``1` `        ``cnt1``-``=``1` `     `  `    ``# Finally all the 2s ` `    ``while` `(cnt2 > ``0``): ` `        ``arr[i] ``=` `2` `        ``i``+``=``1` `        ``cnt2``-``=``1` `     `  `    ``# Prthe sorted array ` `    ``printArr(arr, n) ` ` `  ` `  `# Driver code ` ` `  `arr ``=` `[``0``, ``1``, ``1``, ``0``, ``1``, ``2``, ``1``, ``2``, ``0``, ``0``, ``0``, ``1``] ` `n ``=` `len``(arr) ` ` `  `sortArr(arr, n) ` ` `  `#This code is contributed by shubhamsingh10 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `  `  `class` `GFG { ` `    ``// Utility function to print the contents of an array ` `    ``static` `void` `printArr(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``Console.Write(arr[i] + ``" "``); ` `    ``} ` `     `  `    ``// Function to sort the array of 0s, 1s and 2s ` `    ``static` `void` `sortArr(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``int` `i, cnt0 = 0, cnt1 = 0, cnt2 = 0; ` `     `  `        ``// Count the number of 0s, 1s and 2s in the array ` `        ``for` `(i = 0; i < n; i++) { ` `            ``switch` `(arr[i]) { ` `            ``case` `0: ` `                ``cnt0++; ` `                ``break``; ` `            ``case` `1: ` `                ``cnt1++; ` `                ``break``; ` `            ``case` `2: ` `                ``cnt2++; ` `                ``break``; ` `            ``} ` `        ``} ` `     `  `        ``// Update the array ` `        ``i = 0; ` `     `  `        ``// Store all the 0s in the beginning ` `        ``while` `(cnt0 > 0) { ` `            ``arr[i++] = 0; ` `            ``cnt0--; ` `        ``} ` `     `  `        ``// Then all the 1s ` `        ``while` `(cnt1 > 0) { ` `            ``arr[i++] = 1; ` `            ``cnt1--; ` `        ``} ` `     `  `        ``// Finally all the 2s ` `        ``while` `(cnt2 > 0) { ` `            ``arr[i++] = 2; ` `            ``cnt2--; ` `        ``} ` `     `  `        ``// Print the sorted array ` `        ``printArr(arr, n); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1 }; ` `        ``int` `n = arr.Length; ` `     `  `        ``sortArr(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by shubhamsingh10 `

Output:

```0 0 0 0 0 1 1 1 1 1 2 2
```
• Complexity Analysis:

• Time Complexity: O(n).
Only two traversals of the array is needed.
• Space Complexity: O(1).
As no extra space is required.

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