Segregate 0’s, 1’s and 2’s in an array

March 25, 2023

Segregate 0’s, 1’s and 2’s in an array in C

Here, in this section we will discuss the C program for segregate 0’s, 1’s and 2’s in an array :

Given an array with 0’s, 1’s and 2’s, we are segregating the 0’s, 1’s and 2’s.

INPUT : 0 1 2 0 1 1 1 1 1 0 1 2 1 0 2 2 2 0 1 2 0 1

OUTPUT : 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2

Segregate 0’s, 1's and 2’s in an array in C

Segregating 0’s, 1’s, and 2’s in an array is a problem that involves sorting an array of integers that contains only the values 0, 1, and 2. The goal is to rearrange the array such that all the 0’s are placed before all the 1’s, and all the 1’s are placed before all the 2’s. This problem is also known as the Dutch National Flag problem, as it was originally posed as a metaphor for the three bands of color in the Dutch flag.On this page we will discuss two methods-

Brute Force Method
Dutch National Flag Method

Brute force approach to segregate 0’s, 1’s, and 2’s in an array is to count the number of 0’s, 1’s, and 2’s in the array, and then overwrite the array with the correct number of each value

Algorithm for brute force

Here’s how the algorithm works:

Initialize three counters for 0’s, 1’s, and 2’s to 0.
Loop through the array and count the number of 0’s, 1’s, and 2’s.
Overwrite the array with the correct number of each value, starting with 0’s, then 1’s, then 2’s.

Size of sub-array with max sum

Sub-array with given sum

Triplet that sum to a given value

Segregate 0’s and 1’s in array

Sort elements in array by frequency

Code for brute force method in C

Run

#include <stdio.h>

void segregate (int arr[], int n)
{
  int count_0 = 0, count_1 = 0, count_2 = 0;

  // Step 1: Count the number of 0's, 1's, and 2's
  for (int i = 0; i < n; i++)
    {
      if (arr[i] == 0)
	{
	  count_0++;
	}
      else if (arr[i] == 1)
	{
	  count_1++;
	}
      else if (arr[i] == 2)
	{
	  count_2++;
	}
    }

  // Step 2: Overwrite the array with the correct number of each value
  int i = 0;
  while (count_0 > 0)
    {
      arr[i] = 0;
      count_0--;
      i++;
    }
  while (count_1 > 0)
    {
      arr[i] = 1;
      count_1--;
      i++;
    }
  while (count_2 > 0)
    {
      arr[i] = 2;
      count_2--;
      i++;
    }
}

int main ()
{
  int arr[] = { 2, 0, 1, 2, 0, 1, 0, 2, 1 };
  int n = sizeof (arr) / sizeof (arr[0]);

  printf ("Array: ");
  for (int i = 0; i < n; i++)
    {
      printf ("%d ", arr[i]);
    }
  printf ("\n");


  segregate (arr, n);

  printf ("Segregated array: ");
  for (int i = 0; i < n; i++)
    {
      printf ("%d ", arr[i]);
    }
  printf ("\n");

  return 0;
}

Output

Array: 2 0 1 2 0 1 0 2 1 
Segregated array: 0 0 0 1 1 1 2 2 2

One approach to segregate 0’s, 1’s, and 2’s in an array is to use the Dutch National Flag algorithm. The basic idea of this algorithm is to maintain three pointers to partition the array into three parts: the 0’s, the 1’s, and the 2’s.

Algorithm for dutch national flag algorithm

Here’s how the algorithm works:

Initialize three pointers: low, mid, and high.
- low points to the beginning of the array
- mid points to the beginning of the unsorted part of the array
- high points to the end of the array
Loop through the array from mid to high:
- If the current element is 0, swap it with the element at the low pointer and increment both low and mid pointers.
- If the current element is 1, leave it in place and just increment the mid pointer.
- If the current element is 2, swap it with the element at the high pointer and decrement the high pointer.
Continue this process until the mid pointer crosses over the high pointer.

Code for dutch national flag method in C

Run

#include <stdio.h>

void swap (int *a, int *b)
{
  int temp = *a;
  *a = *b;
  *b = temp;
}

void segregate (int arr[], int n)
{
  int low = 0, mid = 0, high = n - 1;
  while (mid <= high)
    {
      switch (arr[mid])
	{
	case 0:
	  swap (&arr[low], &arr[mid]);
	  low++;
	  mid++;
	  break;
	case 1:
	  mid++;
	  break;
	case 2:
	  swap (&arr[mid], &arr[high]);
	  high--;
	  break;
	}
    }
}

int main ()
{
  int arr[] = { 2, 0, 1, 2, 0, 1, 0, 2, 1 };
  int n = sizeof (arr) / sizeof (arr[0]);
  
  printf ("Array: ");
  for (int i = 0; i < n; i++)
    {
      printf ("%d ", arr[i]);
    }
  printf ("\n");
  

  segregate (arr, n);

  printf ("Segregated array: ");
  for (int i = 0; i < n; i++)
    {
      printf ("%d ", arr[i]);
    }
  printf ("\n");

  return 0;
}

Output

Array: 2 0 1 2 0 1 0 2 1 
Segregated array: 0 0 0 1 1 1 2 2 2

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