250+ TOP MCQs on Binary Search Iterative and Answers [2024]

Data Structure Multiple Choice Questions on “Binary Search Iterative”.

1. What is the advantage of recursive approach than an iterative approach?
a) Consumes less memory
b) Less code and easy to implement
c) Consumes more memory
d) More code has to be written

Answer: b
Clarification: A recursive approach is easier to understand and contains fewer lines of code.

2. Choose the appropriate code that does binary search using recursion.
a)

public static int recursive(int arr[], int low, int high, int key)
{
	int mid = low + (high - low)/2;
	if(arr[mid] == key)
	{
		return mid;
	}
	else if(arr[mid] < key)
	{
		return recursive(arr,mid+1,high,key);
	}
	else
	{
		return recursive(arr,low,mid-1,key);
	}
}

public static int recursive(int arr[], int low, int high, int key)
{
	int mid = low + (high + low)/2;
	if(arr[mid] == key)
	{
		return mid;
	}
	else if(arr[mid] < key)
	{
		return recursive(arr,mid-1,high,key);
	}
	else
	{
		return recursive(arr,low,mid+1,key);
	}
}

public static int recursive(int arr[], int low, int high, int key)
{
	int mid = low + (high - low)/2;
	if(arr[mid] == key)
	{
		return mid;
	}
	else if(arr[mid] < key)
	{
		return recursive(arr,mid,high,key);
	}
	else
	{
		return recursive(arr,low,mid-1,key);
	}
}

public static int recursive(int arr[], int low, int high, int key)
{
	int mid = low + ((high - low)/2)+1;
	if(arr[mid] == key)
	{
		return mid;
	}
	else if(arr[mid] < key)
	{
		return recursive(arr,mid,high,key);
	}
	else
	{
		return recursive(arr,low,mid-1,key);
	}
}

Answer: a
Clarification: Calculate the ‘mid’ value, and check if that is the key, if not, call the function again with 2 sub arrays, one with till mid-1 and the other starting from mid+1.

3. Given an input arr = {2,5,7,99,899}; key = 899; What is the level of recursion?
a) 5
b) 2
c) 3
d) 4

Answer: c
Clarification: level 1: mid = 7
level 2: mid = 99
level 3: mid = 899(this is the key).

4. Given an array arr = {45,77,89,90,94,99,100} and key = 99; what are the mid values(corresponding array elements) in the first and second levels of recursion?
a) 90 and 99
b) 90 and 94
c) 89 and 99
d) 89 and 94

Answer: a
Clarification: At first level key = 90
At second level key= 99
Here 90 and 99 are mid values.

5. What is the worst case complexity of binary search using recursion?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n²)

Answer: b
Clarification: Using the divide and conquer master theorem.

6. What is the average case time complexity of binary search using recursion?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n²)
View Answer

Answer: b
Clarification: T(n) = T(n/2) + 1, Using the divide and conquer master theorem.

7. Which of the following is not an application of binary search?
a) To find the lower/upper bound in an ordered sequence
b) Union of intervals
c) Debugging
d) To search in unordered list
View Answer

Answer: d
Clarification: In Binary search, the elements in the list should be sorted. It is applicable only for ordered list. Hence Binary search in unordered list is not an application.

8. Choose among the following code for an iterative binary search.
a)

public static int iterative(int arr[], int key)
{
	int low = 0;
	int mid = 0;
	int high = arr.length-1;
	while(low <= high)
	{
		mid = low + (high + low)/2;
		if(arr[mid] == key)
		{
			return mid;
		}
		else if(arr[mid] < key)
		{
			low = mid - 1;
		}
		else
		{
			high = mid + 1;
		}
	}
	return -1;
}

public static int iterative(int arr[], int key)
{
	int low = 0;
	int mid = 0;
	int high = arr.length-1;
	while(low <= high)
	{
		mid = low + (high - low)/2;
		if(arr[mid] == key)
		{
			return mid;
		}
		else if(arr[mid] < key)
		{
			low = mid + 1;
		}
		else
		{
			high = mid - 1;
		}
	}
	return -1;
}

public static int iterative(int arr[], int key)
{
	int low = 0;
	int mid = 0;
	int high = arr.length-1;
	while(low <= high)
	{
		mid = low + (high + low)/2;
		if(arr[mid] == key)
		{
			return mid;
		}
		else if(arr[mid] < key)
		{
			low = mid + 1;
		}
		else
		{
			high = mid - 1;
		}
	}
	return -1;
}

public static int iterative(int arr[], int key)
{
	int low = 0;
	int mid = 0;
	int high = arr.length-1;
	while(low <= high)
	{
		mid = low + (high - low)/2;
		if(arr[mid] == key)
		{
			return mid;
		}
		else if(arr[mid] < key)
		{
			low = mid - 1;
		}
		else
		{
			high = mid + 1;
		}
	}
	return -1;
}

Answer: b
Clarification: Find the ‘mid’, check if it equals the key, if not, continue the iterations until low <= high.

9. Binary Search can be categorized into which of the following?
a) Brute Force technique
b) Divide and conquer
c) Greedy algorithm
d) Dynamic programming

Answer: b
Clarification: Since ‘mid’ is calculated for every iteration or recursion, we are diving the array into half and then try to solve the problem.

10. Given an array arr = {5,6,77,88,99} and key = 88; How many iterations are done until the element is found?
a) 1
b) 3
c) 4
d) 2

Answer: d
Clarification: Iteration1 : mid = 77; Iteration2 : mid = 88;

11. Given an array arr = {45,77,89,90,94,99,100} and key = 100; What are the mid values(corresponding array elements) generated in the first and second iterations?
a) 90 and 99
b) 90 and 100
c) 89 and 94
d) 94 and 99

Answer: a
Clarification: Trace the input with the binary search iterative code.

12. What is the time complexity of binary search with iteration?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n²)

Answer: b
Clarification: T(n) = T(n/2) + theta(1)
Using the divide and conquer master theorem, we get the time complexity as O(logn).

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