Given a sorted array of integers, what can be the minimum worst-case time complexity to find ceiling of a number x in given array? The ceiling of an element x is the smallest element present in array which is greater than or equal to x. Ceiling is not present if x is greater than the maximum element present in array. For example, if the given array is {12, 67, 90, 100, 300, 399} and x = 95, then the output should be 100.
(A)
O(loglogn)
(B)
O(n)
(C)
O(log(n))
(D)
O(log(n) * log(n))
Answer: (C)
Explanation:
We modify the standard binary search to find the ceiling. The time complexity T(n) can be written as T(n) <= T(n/2) + O(1) Solution of the above recurrence can be obtained by Master Method. It falls in case 2 of the Master Method. The solution is O(Logn).
int ceilSearch(int arr[], int low, int high, int x) { int mid; /* If x is smaller than or equal to the first element, then return the first element */ if (x <= arr[low]) return low; /* If x is greater than the last element, then return -1 */ if (x > arr[high]) return -1; /* get the index of middle element of arr[low..high]*/ mid = (low + high) / 2; /* low + (high - low)/2 */ /* If x is same as middle element, then return mid */ if (arr[mid] == x) return mid; /* If x is greater than arr[mid], then either arr[mid + 1] is ceiling of x or ceiling lies in arr[mid+1...high] */ else if (arr[mid] < x) { if (mid + 1 <= high && x <= arr[mid + 1]) return mid + 1; else return ceilSearch(arr, mid + 1, high, x); } /* If x is smaller than arr[mid], then either arr[mid] is ceiling of x or ceiling lies in arr[mid-1...high] */ else { if (mid - 1 >= low && x > arr[mid - 1]) return mid; else return ceilSearch(arr, low, mid - 1, x); } }
Hence Option(C) is the correct answer.
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