Frequency In A Sorted Array
You are given a sorted array 'ARR' and a number 'X'. Your task is to count the number of occurrences of 'X' in 'ARR'.

Note :

1. If 'X' is not found in the array, return 0. 2. The given array is sorted in non-decreasing order.

Input Format :

The first line of input contains an integer ‘T’ representing the number of test cases. Then the test cases follow. The first line of each test case contains an integer ‘N’ representing the size of the input array. The second line of each test case contains elements of the array separated by a single space. The last line of each test case contains a single integer 'X'.

Output Format :

For each test case, the only line of output will contain a single integer which represents the number of occurrences of target element 'X' in the array.

Follow Up :

Try to solve it in O(log(N)) time and O(1) space.

Note :

You don't need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1

Question

Frequency In A Sorted Array

You are given a sorted array 'ARR' and a number 'X'. Your task is to count the number of occurrences of 'X' in 'ARR'.

Note :

1. If 'X' is not found in the array, return 0.
2. The given array is sorted in non-decreasing order.

Input Format :

The first line of input contains an integer ‘T’ representing the number of test cases. Then the test cases follow.

The first line of each test case contains an integer ‘N’ representing the size of the input array.

The second line of each test case contains elements of the array separated by a single space.

The last line of each test case contains a single integer 'X'.

Output Format :

For each test case, the only line of output will contain a single integer which represents the number of occurrences of target element 'X' in the array.

Follow Up :

Try to solve it in O(log(N)) time and O(1) space.

Note :

You don't need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1 <= T <= 10^2
1 <= N <= 10^4
0 <= ARR[i], X <= 10^9

Time Limit : 1 sec

Accepted Answer

The task is to count the number of occurrences of a given number in a sorted array.

Use binary search to find the first and last occurrence of the given number in the array.
Subtract the indices of the first and last occurrence to get the count.
Handle the case when the number is not found in the array.

Accepted Answer

1. Gave a map based solution asked me to optimize space and time
2. Gave a linear solution with a counter with no extra space
3. Gave a binary search based solution with upper bound and lower bound. The difference between upper and lower bound will give the frequency.

Note :The interviewer asked me to implement upper and lower bound from scratch and with little to no time left I was unable to implement the upper and lower bound

Accepted Answer

Binary Search

The basic idea is to first find an occurrence of the target element ‘X’ using binary search. Then, we count the matches toward the left and right sides of the found index.

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log(N)).

We basically ignore half of the elements just after one comparison.

Here is the algorithm :

Compare ‘X’ with the middle element.
If ‘X’ matches with the middle element, we return the ‘MID’ index.
Else if ‘X’ is greater than the ‘MID’ element, then ‘X’ can only lie in the right half subarray after the ‘MID’ element. So we recur for the right half.
Else ('X' is smaller) recur for the left half.

Space Complexity: O(logn)Explanation:

O(log(N)), where ‘N’ is the size of the given array.

Recursive stack will be called maximum of log(N) times as we are repeatedly dividing the array into half parts. Hence, the overall space complexity will be O(log(N)).

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the size of the given array.

Time complexities of the functions we have used:

Binary Search : The worst case time complexity of the binary search algorithm is O(log(N)), whereas, the best case time complexity would be O(1) when the central index would directly match the desired value.
To count the occurrences : The worst case time complexity of this function is O(N) when the array contains only the target value x.

Hence, the overall time complexity is O(N + log(N)) i.e. O(N).

Accepted Answer

Linear Search

Linearly search for target element ‘X’, count the occurrences of ‘X’ and return the ‘COUNT’.

Linear search is a method for finding an element within a list/array. It sequentially checks each element of the list until a match is found or the whole list/array has been searched.

Space Complexity: O(1)Explanation:

O(1)

We are not using any extra space. Hence, the overall space complexity will be O(1).

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the size of the given array.

We are traversing the whole array to count the number of occurences. Hence, the overall time complexity will be O(N).

Accepted Answer

Modified Binary Search

The basic idea is to first find the first and the last occurrence of the target element ‘X’ using binary search. Then, return the difference of these indexes +1, as this the count of all the occurrences of the target element ‘X’ in the array.

Here is the algorithm :

Use Binary search to get the index of the first occurrence of target element ‘X’ in ‘ARR’. Let the index of the first occurrence be ‘i’.
Use Binary search to get the index of the last occurrence of target element ‘X’ in ‘ARR’. Let the index of the last occurrence be ‘j’.
Return ('j' - ‘i’ + 1).

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log(N)).

We basically ignore half of the elements just after one comparison.

Here is the algorithm :

Compare ‘X’ with the middle element.
If ‘X’ matches with the middle element, we return the ‘MID’ index.
Else if ‘X’ is greater than the ‘MID’ element, then ‘X’ can only lie in the right half subarray after the mid element. So replace the ‘START’ index by ‘MID’ + 1.
Else ('X' is smaller) replace the ‘END’ index by ‘MID’ - 1.

Space Complexity: O(1)Explanation:

O(1)

We are not using any extra space. Hence, the overall space complexity will be O(1).

Time Complexity: O(logn)Explanation:

O(log(N)), where ‘N’ is the size of the given array.

We repeatedly divide array into half to find the element. Hence, the overall time complexity will be O(log(N)).

You are given a sorted array 'ARR' and a number 'X'. Your task is to count the number of occurrences of 'X' in 'ARR'.

Note :

Input Format :

Output Format :

Follow Up :

Note :

Constraints :

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