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

16 Dec, 2021
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: 
a[1..Lo-1] zeroes (red)a[Lo..Mid-1] ones (white)a[Mid..Hi] unknowna[Hi+1..N] twos (blue)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: 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).Traverse the array from start to end and mid is less than high. (Loop counter is i)If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1If the element is 1 then update mid = mid + 1If 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 processedPrint 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:CC// 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 codeint 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: Time Complexity: O(n). Only one traversal of the array is needed.Space Complexity: O(1). No extra space is required.Please refer complete article on Sort an array of 0s, 1s and 2s for more details!My Personal Notes
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a[1..Lo-1] zeroes (red)
a[Lo..Mid-1] ones (white)
a[Mid..Hi] unknown
a[Hi+1..N] twos (blue)
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: 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).Traverse the array from start to end and mid is less than high. (Loop counter is i)If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1If the element is 1 then update mid = mid + 1If 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 processedPrint 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:CC// 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 codeint 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: Time Complexity: O(n). Only one traversal of the array is needed.Space Complexity: O(1). No extra space is required.Please refer complete article on Sort an array of 0s, 1s and 2s for more details!My Personal Notes
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Algorithm: 
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).Traverse the array from start to end and mid is less than high. (Loop counter is i)If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1If the element is 1 then update mid = mid + 1If 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 processedPrint 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:CC// 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 codeint 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: Time Complexity: O(n). Only one traversal of the array is needed.Space Complexity: O(1). No extra space is required.Please refer complete article on Sort an array of 0s, 1s and 2s for more details!My Personal Notes
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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).
Traverse the array from start to end and mid is less than high. (Loop counter is i)
If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1
If the element is 1 then update mid = mid + 1
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
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:CC// 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 codeint 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: Time Complexity: O(n). Only one traversal of the array is needed.Space Complexity: O(1). No extra space is required.Please refer complete article on Sort an array of 0s, 1s and 2s for more details!My Personal Notes
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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 <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 codeint 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: 
Time Complexity: O(n). Only one traversal of the array is needed.
Space Complexity: O(1). No extra space is required.
Please refer complete article on Sort an array of 0s, 1s and 2s for more details!
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