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

The Big Idea

What if you could count repeated items in a huge sorted list in just a blink?

The Scenario

Imagine you have a long list of sorted numbers, like a phone book sorted by last name. You want to find out how many times a particular name appears. Doing this by checking each entry one by one would take a lot of time.

The Problem

Manually counting each occurrence means going through the entire list, even if the list is very long. This is slow and tiring, and you might lose track or make mistakes while counting.

The Solution

Using a smart method that takes advantage of the list being sorted, we can quickly find where the element starts and ends. This way, we count occurrences fast without checking every item.

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);
    int last = findLast(arr, size, target);
    if (first == -1) return 0;
    return last - first + 1;
}

What It Enables

This method lets you count how many times an element appears in a sorted list very quickly, even if the list is huge.

Real Life Example

When searching a sorted database of customer IDs, you can instantly find how many times a specific ID appears without scanning the entire database.

Key Takeaways

Manual counting is slow and error-prone for large sorted lists.

Using the sorted order, we find the first and last positions of the element quickly.

This approach saves time and reduces mistakes.

Practice

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

easy

A. It reduces the search time from linear to logarithmic

B. It sorts the array automatically

C. It uses less memory than linear search

D. It works only for unsorted arrays

Solution

1. Step 1: Understand linear vs binary search time

   Linear search checks each element one by one, taking O(n) time. Binary search divides the search space in half each time, taking O(log n) time.

2. Step 2: Apply binary search to counting occurrences

   Using binary search to find the first and last occurrence of the element reduces the total search time significantly compared to scanning the whole array.

3. Final Answer:

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

4. Quick Check:

   Binary search = faster search [OK]

Hint: Binary search halves search space, speeding up counting [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++ function signatures is correct for finding the first occurrence of an element in a sorted array using binary search?

easy

A. int firstOccurrence(int arr[], int x)

B. void firstOccurrence(int arr[], int n, int x)

C. int firstOccurrence(int arr, int n, int x)

D. int firstOccurrence(int arr[], int n, int x)

Solution

1. Step 1: Check parameter types and count

   The function needs the array (int arr[]), its size (int n), and the target element (int x).

2. Step 2: Check return type

   It should return an int index of the first occurrence, so return type is int.

3. Final Answer:

   int firstOccurrence(int arr[], int n, int x) -> Option D

4. Quick Check:

   Correct parameters and return type [OK]

Hint: Function returns index, needs array, size, and target [OK]

Common Mistakes:

• Missing array size parameter
• Wrong return type (void instead of int)
• Incorrect array parameter type

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

int arr[] = {1, 2, 2, 2, 3, 4, 5};
int n = 7;
int x = 2;

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

std::cout << countOccurrences(arr, n, x) << std::endl;

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Find first occurrence of 2

   Binary search finds first 2 at index 1.

2. Step 2: Find last occurrence of 2

   Binary search finds last 2 at index 3.

3. Step 3: Calculate count

   Count = last - first + 1 = 3 - 1 + 1 = 3.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   Occurrences of 2 = 3 [OK]

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

Common Mistakes:

• Returning last index instead of count
• Not updating low and high correctly
• Returning 0 when element exists

4. Identify the bug in the following code for counting occurrences of an element in a sorted array:

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

medium

A. Returning last - first instead of last - first + 1

B. Using 'while (low < high)' instead of 'while (low <= high)' causes missing last element

C. Incorrect calculation of mid index

D. Not initializing 'first' and 'last' to zero

Solution

1. Step 1: Analyze loop condition

   The loops use 'while (low < high)', which misses checking when low equals high, possibly skipping the last element.

2. Step 2: Correct loop condition

   Changing to 'while (low <= high)' ensures all elements are checked, including the last one.

3. Final Answer:

   Using 'while (low < high)' instead of 'while (low <= high)' causes missing last element -> Option B

4. Quick Check:

   Loop condition must include equality [OK]

Hint: Use <= in binary search loops to include all elements [OK]

Common Mistakes:

• Using < instead of <= in loops
• Not updating first and last correctly
• Miscomputing mid index

5. Given a sorted array with possible duplicates, how would you modify the binary search approach to count occurrences of multiple different elements efficiently without repeating the search for each element separately?

hard

A. Sort the array again before each search

B. Run binary search separately for each element to find counts

C. Precompute a frequency map using a single pass, then query counts in O(1)

D. Use linear search for each element to count occurrences

Solution

1. Step 1: Understand repeated binary search cost

   Running binary search for each element separately costs O(m log n) for m queries, which can be expensive.

2. Step 2: Use frequency map for efficiency

   A single pass over the sorted array can build a frequency map (element to count) in O(n). Then queries are O(1) each.

3. Final Answer:

   Precompute a frequency map using a single pass, then query counts in O(1) -> Option C

4. Quick Check:

   Frequency map = fast multiple queries [OK]

Hint: Build frequency map once, then answer queries instantly [OK]

Common Mistakes:

• Running binary search repeatedly for each query
• Sorting array multiple times unnecessarily
• Using linear search for each query