Algorithms | Searching | Question 2

Which of the following is correct recurrence for worst case of Binary Search?
(A) T(n) = 2T(n/2) + O(1) and T(1) = T(0) = O(1)
(B) T(n) = T(n-1) + O(1) and T(1) = T(0) = O(1)
(C) T(n) = T(n/2) + O(1) and T(1) = T(0) = O(1)
(D) T(n) = T(n-2) + O(1) and T(1) = T(0) = O(1)


Answer: (C)

Explanation: Following is typical implementation of Binary Search.

// Searches x in arr[low..high].  If x is present, then returns its index, else -1
int binarySearch(int arr[], int low, int high, int x)
{
    if(high >= low)
    {
        int mid = low + (high - low)/2;
        if (x == arr[mid])
            return mid;
        if (x> arr[mid])
            return binarySearch(arr, (mid + 1), high);
        else
            return binarySearch(arr, low, (mid -1));
    }

    return -1;
}

In Binary Search, we first compare the given element x with middle of the array. If x matches with middle element, then we return middle index. Otherwise, we either recur for left half of array or right half of array.

So recurrence is T(n) = T(n/2) + O(1)

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