Question

6. Let T(1..n] be a sorted array of distinct integers, some of which may be negative. Give an algorithm that can find an index i such that 1 <i<n and T[i] = i, provided such an index exists. Your algorithm should take a time in O(lg n) in the worst case. Answers must be proven (or at least well justified)

Answer 1

ALGORITHM:-

1. Compare mid with the middle element.
2. If mid matches with middle element, we return the mid index.
3. Else If mid is greater than the mid element, then mid can only lie in right half subarray after the mid element. So we recur for right half.
4. Else (mid is smaller) recur for the left half.

Calculating Time complexity:

• Hence, the time complexity of Binary Search is

  log₂ (n)

  • Let say the iteration in Binary Search terminates after k iterations. In the above example, it terminates after 3 iterations, so here k = 3
  • At each iteration, the array is divided by half. So let’s say the length of array at any iteration is n
  • At Iteration 1,

    Length of array = n

  • At Iteration 2,

    Length of array = n⁄2

  • At Iteration 3,

    Length of array = (n⁄2)⁄2 = n⁄22

  • Therefore, after Iteration k,

    Length of array = n⁄2k

  • Also, we know that after

    After k divisions, the length of array becomes 1

  • Therefore

    Length of array = n⁄2k = 1
    => n = 2k
    

  • Applying log function on both sides:

    => log2 (n) = log2 (2k)
    => log2 (n) = k log2 (2)
    

  • As (log_a (a) = 1)
    Therefore,

    => k = log2 (n)
    

PROGRAM:-

// C++ program to check fixed point
// in an array using binary search
#include <iostream>
using namespace std;

int binarySearch(int arr[], int low, int high)
{
   if(high >= low)
   {
       int mid = (low + high)/2; /*low + (high - low)/2;*/
       if(mid == arr[mid])
           return mid;
       if(mid > arr[mid])
           return binarySearch(arr, (mid + 1), high);
       else
           return binarySearch(arr, low, (mid -1));
   }

   /* Return -1 if there is no Fixed Point */
   return -1;
}

/* Driver code */
int main()
{
   int arr[10] = {-10, -1, 0, 3, 10, 11, 30, 50, 100};
   int n = sizeof(arr)/sizeof(arr[0]);
   cout<<"Fixed Point is "<< binarySearch(arr, 0, n-1);
   return 0;
}

OUTPUT:-

Fixed Point is 3