Count Occurrences of Element in Sorted Array in DSA C# - Time & Space Complexity

Time Complexity: Count Occurrences of Element in Sorted Array

O(log n)

Understanding Time Complexity

We want to know how long it takes to count how many times a number appears in a sorted list.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int CountOccurrences(int[] arr, int target) {
    int first = FindFirst(arr, target);
    if (first == -1) return 0;
    int last = FindLast(arr, target);
    return last - first + 1;
}

int FindFirst(int[] arr, int target) {
    int low = 0, high = arr.Length - 1;
    int result = -1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == target) {
            result = mid;
            high = mid - 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

int FindLast(int[] arr, int target) {
    int low = 0, high = arr.Length - 1;
    int result = -1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == target) {
            result = mid;
            low = mid + 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}
    

This code finds the first and last positions of the target number using binary search, then calculates how many times it appears.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Binary search loops in FindFirst and FindLast methods.
• How many times: Each binary search loop runs about log₂(n) times, where n is the array size.

How Execution Grows With Input

Each binary search cuts the search space in half each time, so the number of steps grows slowly as the list grows.

Input Size (n)	Approx. Operations
10	About 7 (2 x log₂(10))
100	About 14 (2 x log₂(100))
1000	About 20 (2 x log₂(1000))

Pattern observation: Operations grow slowly, roughly doubling the input size adds only a few more steps.

Final Time Complexity

Time Complexity: O(log n)

This means the time to count occurrences grows slowly as the list gets bigger, because we use fast searching.

Common Mistake

[X] Wrong: "Counting occurrences requires checking every element, so it must be O(n)."

[OK] Correct: Because the array is sorted, we can find the first and last positions quickly using binary search, avoiding checking every element.

Interview Connect

Knowing how to use binary search to count occurrences shows you can use the sorted property to make searching efficient, a skill often valued in interviews.

Self-Check

"What if the array was not sorted? How would the time complexity change?"

Practice

1. What is the main advantage of using a binary search approach to count occurrences of an element in a sorted array?

easy

A. It works only for unsorted arrays.

B. It uses less memory than a linear scan.

C. It reduces the time complexity to O(log n) compared to a linear scan.

D. It counts occurrences without checking the array elements.

Solution

1. Step 1: Understand binary search efficiency

   Binary search divides the search space in half each time, making it faster than checking every element.

2. Step 2: Compare with linear scan

   Linear scan checks each element, taking O(n) time, while binary search takes O(log n) time.

3. Final Answer:

   It reduces the time complexity to O(log n) compared to a linear scan. -> Option C

4. Quick Check:

   Binary search = O(log n) speed [OK]

Hint: Binary search halves search space, speeding up counting [OK]

Common Mistakes:

• Thinking binary search works on unsorted arrays
• Confusing time complexity with memory usage
• Assuming binary search counts without element checks

2. Which of the following C# method signatures correctly represents a function to count occurrences of an integer in a sorted array?

easy

A. bool CountOccurrences(int[] arr, int target)

B. int CountOccurrences(int[] arr, int target)

C. int CountOccurrences(int target, int[] arr)

D. void CountOccurrences(int[] arr, int target)

Solution

1. Step 1: Identify return type for count

   The function should return an integer representing the count, so return type must be int.

2. Step 2: Check parameter order and types

   Common convention is array first, then target element; parameters types must match int array and int target.

3. Final Answer:

   int CountOccurrences(int[] arr, int target) -> Option B

4. Quick Check:

   Return count as int, array then target [OK]

Hint: Count returns int; array parameter comes before target [OK]

Common Mistakes:

• Using void return type instead of int
• Swapping parameter order incorrectly
• Returning bool instead of count

3. What will be the output of the following C# code snippet?

int[] arr = {1, 2, 2, 2, 3, 4, 5};
int target = 2;
int count = CountOccurrences(arr, target);
Console.WriteLine(count);

int CountOccurrences(int[] nums, int x) {
    int first = Array.BinarySearch(nums, x);
    if (first < 0) return 0;
    int count = 1;
    int left = first - 1;
    while (left >= 0 && nums[left] == x) {
        count++;
        left--;
    }
    int right = first + 1;
    while (right < nums.Length && nums[right] == x) {
        count++;
        right++;
    }
    return count;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Find first occurrence index using BinarySearch

   BinarySearch finds one occurrence of 2 at index 2.

2. Step 2: Count occurrences left and right of index 2

   Left side: index 1 and 0 checked; index 1 is 2 (count=2), index 0 is 1 (stop). Right side: index 3 is 2 (count=3), index 4 is 3 (stop).

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Occurrences of 2 = 3 [OK]

Hint: BinarySearch finds one index; count neighbors equal to target [OK]

Common Mistakes:

• Assuming BinarySearch returns first occurrence always
• Counting only one side neighbors
• Returning 0 if not found without checking neighbors

4. Identify the bug in the following C# method that counts occurrences of a target in a sorted array:

int CountOccurrences(int[] nums, int target) {
    int first = Array.BinarySearch(nums, target);
    if (first < 0) return 0;
    int count = 1;
    int left = first - 1;
    while (left > 0 && nums[left] == target) {
        count++;
        left--;
    }
    int right = first + 1;
    while (right < nums.Length && nums[right] == target) {
        count++;
        right++;
    }
    return count;
}

medium

A. The method should return -1 if target not found instead of 0.

B. The right boundary check should be 'right <= nums.Length' instead of 'right < nums.Length'.

C. The count variable should start at 0, not 1.

D. The left boundary check should be 'left >= 0' instead of 'left > 0'.

Solution

1. Step 1: Analyze left boundary condition in while loop

   The loop checks 'left > 0', so it skips index 0, missing a possible occurrence at start.

2. Step 2: Correct boundary condition to include index 0

   Changing condition to 'left >= 0' ensures index 0 is checked for target equality.

3. Final Answer:

   The left boundary check should be 'left >= 0' instead of 'left > 0'. -> Option D

4. Quick Check:

   Check array start index 0 with '>= 0' [OK]

Hint: Use '>= 0' to include first array index in left scan [OK]

Common Mistakes:

• Using 'left > 0' misses first element
• Returning -1 instead of 0 when not found
• Starting count at 0 instead of 1
• Using wrong right boundary condition

5. Given a sorted array with possible duplicates, how can you efficiently count occurrences of a target element using two binary searches in C#?

hard

A. Find the first and last occurrence indices of the target using binary search, then calculate count as last - first + 1.

B. Use a single binary search to find any occurrence, then count neighbors linearly.

C. Sort the array again and count occurrences with a linear scan.

D. Use a hash map to store counts of all elements, then retrieve target count.

Solution

1. Step 1: Use binary search to find first occurrence of target

   Modify binary search to find the lowest index where target appears.

2. Step 2: Use binary search to find last occurrence of target

   Modify binary search to find the highest index where target appears.

3. Step 3: Calculate count from indices

   Count = last occurrence index - first occurrence index + 1.

4. Final Answer:

   Find the first and last occurrence indices of the target using binary search, then calculate count as last - first + 1. -> Option A

5. Quick Check:

   Count = last - first + 1 using two binary searches [OK]

Hint: Two binary searches find boundaries; subtract indices for count [OK]

Common Mistakes:

• Using linear scan after one binary search
• Sorting array again unnecessarily
• Using hash map instead of binary search