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Sort an array of 0s, 1s and 2s
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/* | |
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. | |
Example | |
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} | |
soln:Dutch National Flag Algorithm, or 3-way Partitioning — | |
*/ | |
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; | |
} | |
} | |
} |
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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 put following loop before switch condition to make sure that hi is pointing to location which is not 2 so that it would eliminate unnecessary swaps of 2.
while ((a[hi]==2) && hi >= mid)
–hi;
if (hi < mid)
break;