Why Count Occurrences of Element in Sorted Array in DSA C?

The Big Idea

What if you could count repeated numbers in a huge list without checking each one?

The Scenario

Imagine you have a long list of sorted numbers, like a phone book sorted by phone numbers, and you want to find out how many times a specific number appears.

Doing this by checking each number one by one would take a lot of time, especially if the list is very long.

The Problem

Manually scanning through the entire list to count occurrences is slow and tiring.

It's easy to make mistakes by missing some numbers or counting wrong.

This approach wastes time and effort, especially when the list is huge.

The Solution

Using a smart method that takes advantage of the list being sorted, we can quickly find where the number starts and ends.

This way, we count the occurrences without checking every single number.

This saves time and reduces errors.

Before vs After

✗ Before

int countOccurrences(int arr[], int size, int target) {
    int count = 0;
    for (int i = 0; i < size; i++) {
        if (arr[i] == target) count++;
    }
    return count;
}

✓ After

int countOccurrences(int arr[], int size, int target) {
    int first = findFirst(arr, size, target);
    if (first == -1) return 0;
    int last = findLast(arr, size, target);
    return last - first + 1;
}

What It Enables

This method lets you quickly and accurately count how many times a number appears in a sorted list, even if the list is very large.

Real Life Example

Counting how many times a specific product ID appears in a sorted sales record to analyze its popularity.

Key Takeaways

Manual counting is slow and error-prone.

Using the sorted order helps find occurrences quickly.

This method saves time and improves accuracy.

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 uses less memory than linear search

B. It sorts the array automatically

C. It reduces the search time from linear to logarithmic

D. It works only on unsorted arrays

Solution

1. Step 1: Understand binary search efficiency

   Binary search divides the search space in half each time, reducing time complexity to O(log n).

2. Step 2: Compare with linear search

   Linear search checks each element one by one, taking O(n) time, which is slower for large arrays.

3. Final Answer:

   It reduces the search time from linear to logarithmic -> Option C

4. Quick Check:

   Binary search = faster search [OK]

Hint: Binary search halves search space each step [OK]

Common Mistakes:

• Thinking binary search sorts the array
• Confusing memory usage with speed
• Assuming binary search works on unsorted arrays

2. Which of the following C code snippets correctly declares a function to count occurrences of an integer in a sorted array?

easy

A. void countOccurrences(int arr[], int n, int x);

B. int countOccurrences(int arr[], int n, int x);

C. int countOccurrences(int arr, int n, int x);

D. int countOccurrences(int arr[], int x);

Solution

1. Step 1: Check function return type and parameters

   The function should return an int (count), take array, size, and target element as parameters.

2. Step 2: Validate parameter types

   Array parameter must be int arr[] or int *arr, size as int n, and target as int x.

3. Final Answer:

   int countOccurrences(int arr[], int n, int x); -> Option B

4. Quick Check:

   Correct function signature = int countOccurrences(int arr[], int n, int x); [OK]

Hint: Function returns count as int, array passed as int[] [OK]

Common Mistakes:

• Using void return type instead of int
• Passing array as int instead of int[]
• Missing array size parameter

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

int arr[] = {1, 2, 2, 2, 3, 4, 5};
int n = 7;
int x = 2;
int count = 0;
for (int i = 0; i < n; i++) {
    if (arr[i] == x) count++;
}
printf("%d", count);

medium

A. 1

B. 2

C. 4

D. 3

Solution

1. Step 1: Iterate through the array and count matches

   Elements equal to 2 are at indices 1, 2, and 3, so count increments 3 times.

2. Step 2: Print the count

   The final count is 3, so printf outputs 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Count of 2 in array = 3 [OK]

Hint: Count elements equal to x by scanning array once [OK]

Common Mistakes:

• Counting only first occurrence
• Off-by-one errors in loop
• Printing count before loop ends

4. Identify the error in the following C function intended to count occurrences of x in a sorted array using binary search:

int countOccurrences(int arr[], int n, int x) {
    int count = 0;
    int low = 0, high = n - 1;
    while (low <= high) {
        int mid = (low + high) / 2;
        if (arr[mid] == x) {
            count++;
            mid++;
        } else if (arr[mid] < x) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return count;
}

medium

A. Incrementing mid inside the loop does not change low or high

B. The loop condition should be low < high

C. count should be initialized to -1

D. The function should return void

Solution

1. Step 1: Analyze loop variables and updates

   Incrementing mid does not affect low or high, so loop may become infinite or incorrect.

2. Step 2: Understand binary search update rules

   To move search window, low or high must be updated, not mid.

3. Final Answer:

   Incrementing mid inside the loop does not change low or high -> Option A

4. Quick Check:

   mid++ does not move search window [OK]

Hint: Update low/high, not mid, to move binary search window [OK]

Common Mistakes:

• Changing mid instead of low/high
• Wrong loop condition causing infinite loop
• Incorrect return type

5. Given a sorted array arr = {1, 2, 2, 2, 3, 3, 4, 5, 5, 5, 5}, write a function that uses binary search to count occurrences of x = 5. What is the count returned?

hard

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Use binary search to find first occurrence of 5

   First 5 is at index 7.

2. Step 2: Use binary search to find last occurrence of 5

   Last 5 is at index 10.

3. Step 3: Calculate count as last - first + 1

   Count = 10 - 7 + 1 = 4.

4. Final Answer:

   4 -> Option A

5. Quick Check:

   Count of 5 in array = 4 [OK]

Hint: Count = last index - first index + 1 using binary search [OK]

Common Mistakes:

• Counting total elements instead of occurrences
• Off-by-one errors in count calculation
• Using linear search instead of binary search