Question

000111: Find the first occurrence of 1 in the best way possible.

This was all the information and details I was given for the question.

Answer 1

Answer:

• The input given is {0,0,0,1,1,1}(lets say an array with 0,1,2,3,4,5 indexes) out of which we have to findout the index of the first occurence of 1.
• When you search linearly from the left to right, it takes 4 comparisons to find the 1.
• You can obeserve that the list is already sorted. So we can apply binary search insted of linear search.
• Binary search finds out the middle element first then it compares the required element with middle element.
• Here, middle element will be (0+5)/2=2 where, 5 is the total length of the input[0th element, 1st element,......,5th element]. middle element=(lower limit+upper limit)/2
• So it compares 2nd element with 1.In this case actually the 2nd element is 0.
• So apply binary search on right elements of 2nd elements {1,1,1} to which the upper limit= 5 and lowe limit = 3, middle element is (5+3)/2=4. And the fourth element is 1. But there is possiblity that it might not be the first 1.
• Then take 4th element as upper limit, and 3rd element as lower limit then the middle element will be 4+2/2=3. And which is 1 the first occurace of 1.
• So if you use binary search it take only 3 comparision to get there.
• In a larger view/ input linear search takes n comparisions where as binary takes logn comparisons.

Implementation:

Code:

#include <stdio.h>
int code1(int input[], int ll, int ul);
int main()
{
int input[] = { 0, 0, 0, 1, 1, 1};
   int n = sizeof(input)/sizeof(input[0]);
   printf("%d",code1(input,0,n - 1));
   return 0;
}
int code1(int input[], int ll, int ul)//ll=lower limit, ul=upperlimit
{
   while (ll <= ul) {
       int middle=(ll+ul)/2;
       if (input[middle]==1&&(middle==0||input[middle-1]==0))
           return middle;
       else if (input[middle]==1)
           ul=middle-1;
       else
           ll=middle+1;
   }
   return -1;
}

Output: