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C Program For Sorting An Array Of 0s, 1s and 2s

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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}

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

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)
  5. If the ith element is 0 then swap the element to the low range, thus shrinking the unknown range.
  6. Similarly, if the element is 1 then keep it as it is but shrink the unknown range.
  7. 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]:

      DNF1

      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++
       

      DNF2

      Case (1) a[Mid] is white, Mid++

      DNF3

      Case (2) a[Mid] is blue, swap a[Mid] and a[Hi]; Hi–

      DNF4

      Continue until Mid>Hi.

      Implementation:

      C




      // 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]) 
              {
                  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 code
      int main()
      {
          int arr[] = {0, 1, 1, 0, 1, 2, 
                       1, 2, 0, 0, 0, 1};
          int arr_size = sizeof(arr) / sizeof(arr[0]);
          int i;
        
          sort012(arr, arr_size);
          printf("array after segregation ");
          printArray(arr, arr_size);
          getchar();
          return 0;
      }

      
      

      Output: 

      array after segregation
       0 0 0 0 0 1 1 1 1 1 2 2 

      Complexity Analysis: 

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