quick sort best case code example

Example 1: quick sort program in c

#include<stdio.h>
void quicksort(int number[25],int first,int last){
   int i, j, pivot, temp;

   if(first<last){
      pivot=first;
      i=first;
      j=last;

      while(i<j){
         while(number[i]<=number[pivot]&&i<last)
            i++;
         while(number[j]>number[pivot])
            j--;
         if(i<j){
            temp=number[i];
            number[i]=number[j];
            number[j]=temp;
         }
      }

      temp=number[pivot];
      number[pivot]=number[j];
      number[j]=temp;
      quicksort(number,first,j-1);
      quicksort(number,j+1,last);

   }
}

int main(){
   int i, count, number[25];

   printf("How many elements are u going to enter?: ");
   scanf("%d",&count);

   printf("Enter %d elements: ", count);
   for(i=0;i<count;i++)
      scanf("%d",&number[i]);

   quicksort(number,0,count-1);

   printf("Order of Sorted elements: ");
   for(i=0;i<count;i++)
      printf(" %d",number[i]);

   return 0;
}

Example 2: quicksort

// @see https://www.youtube.com/watch?v=es2T6KY45cA&vl=en
// @see https://www.youtube.com/watch?v=aXXWXz5rF64
// @see https://www.cs.usfca.edu/~galles/visualization/ComparisonSort.html

function partition(list, start, end) {
    const pivot = list[end];
    let i = start;
    for (let j = start; j < end; j += 1) {
        if (list[j] <= pivot) {
            [list[j], list[i]] = [list[i], list[j]];
            i++;
        }
    }
    [list[i], list[end]] = [list[end], list[i]];
    return i;
}

function quicksort(list, start = 0, end = undefined) {
    if (end === undefined) {
        end = list.length - 1;
    }
    if (start < end) {
        const p = partition(list, start, end);
        quicksort(list, start, p - 1);
        quicksort(list, p + 1, end);
    }
    return list;
}

quicksort([5, 4, 2, 6, 10, 8, 7, 1, 0]);

Example 3: analysis of quick sort

T(n) = 2*T(n/2) + n                        // T(n/2) = 2*T(n/4) + (n/2)    

        = 2*[ 2*T(n/4) + n/2 ] + n
	= 22*T(n/4) + n + n
	= 22*T(n/4) + 2n                       // T(n/4) = 2*T(n/8) + (n/4)

	= 22*[ 2*T(n/8) + (n/4) ] + 2n
	= 23*T(n/8) + 22*(n/4) + 2n
	= 23*T(n/8) + n + 2n
	= 23*T(n/8) + 3n

	= 24*T(n/16) + 4n
	and so on....

	= 2k*T(n/(2k)) + k*n      // Keep going until: n/(2k) = 1  <==> n = 2k    

	= 2k*T(1) + k*n
	= 2k*1 + k*n
	= 2k + k*n               // n = 2k
	= n + k*n
	= n + (lg(n))*n
        = n*( lg(n) + 1 )
       ~= n*lg(n))