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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? 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 output should be 100.
- a)O(LogLogn)
- b)O(n)
- c)O(Logn)
- d)O(Logn * Logn)
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Given a sorted array of integers, what can be the minimum worst case t...
Given that the array is sorted, the minimum worst-case time complexity to find the ceiling of a number x in the given array can be achieved using binary search. Binary search operates efficiently on sorted arrays, allowing us to quickly narrow down the range in which the ceiling might be found.
The time complexity of binary search is O(log n).
Thus, the correct answer is: O(Logn)
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Given a sorted array of integers, what can be the minimum worst case t...
Minimum Worst Case Time Complexity to Find Ceiling of a Number x in a Sorted Array
Finding the ceiling of a number x in a sorted array refers to finding the smallest element in the array that is greater than or equal to x. The minimum worst case time complexity for this operation is O(Logn), where n is the number of elements in the array.
Explanation:
To understand why the minimum worst case time complexity is O(Logn), let's consider the steps involved in finding the ceiling of a number x in a sorted array.
1. Initialize two variables, 'low' and 'high', to the first and last indices of the array respectively.
2. While 'low' is less than or equal to 'high', do the following:
- Calculate the middle index as (low + high) / 2.
- If the middle element is less than x, update 'low' to (middle + 1) to search in the right half of the array.
- Otherwise, update 'high' to (middle - 1) to search in the left half of the array.
3. After the loop terminates, the value of 'low' will be pointing to the smallest element greater than or equal to x, if it exists. If 'low' is pointing to the last index of the array, it means the ceiling does not exist.
Key Points:
- In each iteration of the binary search, the search space is divided into half.
- Therefore, if the array has n elements, the maximum number of iterations required to find the ceiling will be Logn.
- Each iteration involves a constant amount of operations, such as calculating the middle index and updating the 'low' or 'high' variables.
- Hence, the time complexity of finding the ceiling of a number x in a sorted array is O(Logn).
Conclusion:
The minimum worst case time complexity to find the ceiling of a number x in a sorted array is O(Logn). This is achieved by using a binary search algorithm, which divides the search space in half at each iteration.
Finding the ceiling of a number x in a sorted array refers to finding the smallest element in the array that is greater than or equal to x. The minimum worst case time complexity for this operation is O(Logn), where n is the number of elements in the array.
Explanation:
To understand why the minimum worst case time complexity is O(Logn), let's consider the steps involved in finding the ceiling of a number x in a sorted array.
1. Initialize two variables, 'low' and 'high', to the first and last indices of the array respectively.
2. While 'low' is less than or equal to 'high', do the following:
- Calculate the middle index as (low + high) / 2.
- If the middle element is less than x, update 'low' to (middle + 1) to search in the right half of the array.
- Otherwise, update 'high' to (middle - 1) to search in the left half of the array.
3. After the loop terminates, the value of 'low' will be pointing to the smallest element greater than or equal to x, if it exists. If 'low' is pointing to the last index of the array, it means the ceiling does not exist.
Key Points:
- In each iteration of the binary search, the search space is divided into half.
- Therefore, if the array has n elements, the maximum number of iterations required to find the ceiling will be Logn.
- Each iteration involves a constant amount of operations, such as calculating the middle index and updating the 'low' or 'high' variables.
- Hence, the time complexity of finding the ceiling of a number x in a sorted array is O(Logn).
Conclusion:
The minimum worst case time complexity to find the ceiling of a number x in a sorted array is O(Logn). This is achieved by using a binary search algorithm, which divides the search space in half at each iteration.
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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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer?.
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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer?.
Solutions for 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? 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 output should be 100.a)O(LogLogn)b)O(n)c)O(Logn)d)O(Logn * Logn)Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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