Sort an array of 0s, 1s and 2s - GeeksforGeeks
Given an array A[] consisting 0s, 1s and 2s, write a function that sorts A[]. The functions should put all 0s first, then all 1s and all 2s in last.
{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}
The problem is similar to our old post , and both of these problems are variation of famous .
The problem was posed with three colours, here `0′, `1′ and `2′. The array is divided into four sections:
a[1..Lo-1] zeroes (red)
a[Lo..Mid-] ones (white)
a[Mid..Hi] unknown
a[Hi+1..N] twos (blue)
The unknown region is shrunk while maintaining these conditions
Lo := 1; Mid := 1; Hi := N;
while Mid &= Hi do
Invariant: a[1..Lo-1]=0 and a[Lo..Mid-1]=1 and a[Hi+1..N]=2;
a[Mid..Hi] are unknown.
case a[Mid] in
0: swap a[Lo] and a[Mid]; Lo++; Mid++
2: swap a[Mid] and a[Hi]; Hi–
— Dutch National Flag Algorithm, or 3-way Partitioning —
[YOU NEED A JavaScript ENABLED BOWSER!!!]
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.
Below is C implementation of above algorithm.
// C program to sort an array with 0,1 and 2
// in a single pass
#include&stdio.h&
/* 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])
swap(&a[lo++], &a[mid++]);
swap(&a[mid], &a[hi--]);
/* UTILITY FUNCTIONS */
void swap(int *a, int *b)
int temp = *a;
/* Utility function to print array arr[] */
void printArray(int arr[], int arr_size)
for (i = 0; i & arr_ 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[0]);
sort012(arr, arr_size);
printf(&array after segregation &);
printArray(arr, arr_size);
getchar();
// 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])
temp = a[mid];
a[mid] = a[hi];
/* Utility function to print array arr[] */
static void printArray(int arr[], int arr_size)
for (i = 0; i & arr_ 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.
sort012(arr, arr_size);
System.out.println(&Array after seggregation &);
printArray(arr, arr_size);
/*This code is contributed by Devesh Agrawal*/
array after segregation 0 0 0 0 0 1 1 1 1 1 2 2
Time Complexity: O(n)
The above code performs unnecessary swaps for inputs like 0 0 0 0 1 1 1 2 2 2 2 2 : lo=4 and mid=7 and hi=11. In present code: first 7 exchanged with 11 and hi become 10 and mid is still pointing to 7. again same operation is till the mid <= hi. But it is really not required. We can change the swap function to do a check that the values being swapped are same or not, if not same, then only swap values.
Thanks to Ankur Roy for suggesting this optimization.
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
Writing code in comment? Please use , generate link and share the link here.
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