Count Occurrences of Element in Sorted Array in DSA C# - Build from Scratch

Count Occurrences of Element in Sorted Array

📖 Scenario: You have a sorted list of product IDs scanned at a store checkout. You want to find out how many times a specific product ID appears in the list.

🎯 Goal: Build a program that counts how many times a given product ID appears in a sorted array of product IDs.

📋 What You'll Learn

Create an integer array called productIds with the exact values: 101, 102, 102, 102, 103, 104, 104, 105

Create an integer variable called targetId and set it to 102

Use a for loop with variable i to iterate over productIds and count how many times targetId appears

Create an integer variable called count to keep track of the number of occurrences

Print the count using Console.WriteLine in the format: "Count of 102: 3"

💡 Why This Matters

🌍 Real World

Counting how many times a product appears in a list helps stores track inventory and sales.

💼 Career

Understanding arrays and loops is fundamental for software developers working with data processing and analysis.

1

Create the sorted array of product IDs

Create an integer array called productIds with these exact values: 101, 102, 102, 102, 103, 104, 104, 105

DSA C#

# Your code here

Hint

Use int[] productIds = { ... }; to create the array with the exact numbers.

2

Set the target product ID to count

Create an integer variable called targetId and set it to 102

DSA C#

int[] productIds = {101, 102, 102, 102, 103, 104, 104, 105};
# Your code here

Hint

Use int targetId = 102; to set the product ID you want to count.

3

Count the occurrences of the target product ID

Create an integer variable called count and set it to 0. Then use a for loop with variable i to iterate over productIds. Inside the loop, increase count by 1 each time productIds[i] equals targetId.

DSA C#

int[] productIds = {101, 102, 102, 102, 103, 104, 104, 105};
int targetId = 102;
# Your code here

Hint

Start count at 0. Use a for loop from 0 to productIds.Length. Inside, check if productIds[i] equals targetId. If yes, add 1 to count.

4

Print the count of the target product ID

Use Console.WriteLine to print the count in this exact format: "Count of 102: 3"

DSA C#

int[] productIds = {101, 102, 102, 102, 103, 104, 104, 105};
int targetId = 102;
int count = 0;
for (int i = 0; i < productIds.Length; i++)
{
    if (productIds[i] == targetId)
    {
        count++;
    }
}
# Your code here

Hint

Use Console.WriteLine($"Count of {targetId}: {count}"); to print the result exactly.

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