Segregate 0’s 1’s and 2’s in an Array

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

On this page we will discuss the C program for segregate 0’s 1’s and 2’s in an array in Java. We are given an array  containing 0,1 and 2 and we have arrange them 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. We will discuss how to solve this problem using different methods along with algorithm and code in Java.

Segregate 0’s and 1’s in an array

Segregate 0’s, 1's and 2’s in an Array in Java

The task at hand is to sort an array of integers that exclusively consists of 0’s, 1’s, and 2’s, with the goal of segregating them in a specific order. The desired arrangement entails placing all the 0’s before the 1’s, and all the 1’s before the 2’s. This particular challenge is commonly known as the Dutch National Flag problem, drawing inspiration from the three bands of color in the Dutch flag. In this discourse, we will look  into two different approaches to tackle this problem.

  1. Brute Force Method
  2. Dutch National Flag Method
Segregate-0’s_-1_s-and-2’s-in-an-array-in-Java

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:

  1. Initialize counters for 0’s, 1’s, and 2’s to 0.
  2. Loop through the array and count the occurrences of 0’s, 1’s, and 2’s.
  3. Store the counts in separate variables.
  4. Overwrite the array with the correct number of each value, starting with 0’s, then 1’s, then 2’s, using the stored counts.

Code for brute force method in Java

Run
import java.util.Arrays;

public class Main {
    public static 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++;
        }
    }

    public static void main(String[] args) {
        int arr[] = {2, 0, 1, 2, 0, 1, 0, 2, 1};
        int n = arr.length;

        System.out.print("Array: ");
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();

        segregate(arr, n);

        System.out.print("Segregated array: ");
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
}

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:

  1. Initialize three pointers – low pointing to the beginning of the array, mid pointing to the current element being processed, and high pointing to the end of the array.
  2. While mid is less than or equal to high, repeat steps 3-7.
  3. Check if the value at arr[mid] is 0.
  4. If yes, swap the values at arr[mid] and arr[low], and increment both mid and low.
  5. If the value at arr[mid] is 1, move mid pointer to the next element.
  6. If the value at arr[mid] is 2, swap the values at arr[mid] and arr[high], and decrement high.
  7. Repeat steps 3-6 until mid becomes greater than high.
  8. The array is now segregated with 0’s on the left, 1’s in the middle, and 2’s on the right.

Code for dutch national flag method in Java

Run
import java.util.Arrays;

public class Main {
    
    public static void dutchNationalFlag(int[] arr, int n) {
        int low = 0;  // pointer to the low end of the array
        int mid = 0;  // pointer to the current element being processed
        int high = n - 1;  // pointer to the high end of the array
        
        while (mid <= high) {
            if (arr[mid] == 0) {
                // Swap arr[mid] and arr[low], increment both mid and low
                int temp = arr[mid];
                arr[mid] = arr[low];
                arr[low] = temp;
                mid++;
                low++;
            } else if (arr[mid] == 1) {
                // Move mid pointer to the next element
                mid++;
            } else if (arr[mid] == 2) {
                // Swap arr[mid] and arr[high], decrement high
                int temp = arr[mid];
                arr[mid] = arr[high];
                arr[high] = temp;
                high--;
            }
        }
    }

    public static void main(String[] args) {
        int[] arr = {2, 0, 1, 2, 0, 1, 0, 2, 1};
        int n = arr.length;

        System.out.print("Array: ");
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();

        dutchNationalFlag(arr, n);

        System.out.print("Segregated array: ");
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
}

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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