Count Occurrences of Element in Sorted Array in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of count occurrences function with duplicates

What is the output of the following C++ code that counts occurrences of 3 in a sorted array?

DSA C++

#include <iostream>
#include <vector>
#include <algorithm>

int countOccurrences(const std::vector<int>& arr, int x) {
    auto first = std::lower_bound(arr.begin(), arr.end(), x);
    auto last = std::upper_bound(arr.begin(), arr.end(), x);
    return last - first;
}

int main() {
    std::vector<int> data = {1, 2, 3, 3, 3, 4, 5};
    std::cout << countOccurrences(data, 3) << std::endl;
    return 0;
}

A 2

B 3

C 1

D 0

💡 Hint

Use lower_bound and upper_bound to find the range of the element.

The element 3 appears three times consecutively in the sorted array. lower_bound finds the first 3, upper_bound finds the position after the last 3, so the difference is 3.

❓ Predict Output

intermediate

Count occurrences of element not present in array

What is the output of this C++ code counting occurrences of 6 in a sorted array?

DSA C++

#include <iostream>
#include <vector>
#include <algorithm>

int countOccurrences(const std::vector<int>& arr, int x) {
    auto first = std::lower_bound(arr.begin(), arr.end(), x);
    auto last = std::upper_bound(arr.begin(), arr.end(), x);
    return last - first;
}

int main() {
    std::vector<int> data = {1, 2, 3, 4, 5};
    std::cout << countOccurrences(data, 6) << std::endl;
    return 0;
}

A 5

B 1

C -1

D 0

💡 Hint

If element is not found, lower_bound and upper_bound will be equal.

Since 6 is not in the array, lower_bound and upper_bound point to the same position, so the count is zero.

🔧 Debug

advanced

Identify the error in counting occurrences code

What error does the following C++ code produce when counting occurrences of 3 in a sorted array?

DSA C++

#include <iostream>
#include <vector>
#include <algorithm>

int countOccurrences(const std::vector<int>& arr, int x) {
    auto first = std::lower_bound(arr.begin(), arr.end(), x);
    auto last = std::upper_bound(arr.begin(), arr.end(), x);
    return first - last;
}

int main() {
    std::vector<int> data = {1, 2, 3, 3, 3, 4, 5};
    std::cout << countOccurrences(data, 3) << std::endl;
    return 0;
}

A Negative number output

B SyntaxError

C Compilation error due to invalid subtraction

D Zero output

💡 Hint

Check the order of subtraction between iterators.

The code subtracts last - first in reverse order, resulting in a negative count.

❓ Predict Output

advanced

Count occurrences with all elements same

What is the output of this C++ code counting occurrences of 7 in an array where all elements are 7?

DSA C++

#include <iostream>
#include <vector>
#include <algorithm>

int countOccurrences(const std::vector<int>& arr, int x) {
    auto first = std::lower_bound(arr.begin(), arr.end(), x);
    auto last = std::upper_bound(arr.begin(), arr.end(), x);
    return last - first;
}

int main() {
    std::vector<int> data(10, 7);
    std::cout << countOccurrences(data, 7) << std::endl;
    return 0;
}

A 10

B 0

C 1

D 9

💡 Hint

All elements are the target value.

Since all 10 elements are 7, the count is 10.

🧠 Conceptual

expert

Why use binary search for counting occurrences in sorted array?

Why is binary search preferred over linear search to count occurrences of an element in a sorted array?

A Binary search can count occurrences without checking elements

B Binary search uses less memory than linear search

C Binary search reduces time complexity from O(n) to O(log n) by exploiting sorted order

D Binary search works only on unsorted arrays

💡 Hint

Think about how sorted order helps search faster.

Binary search quickly finds the range of the element in O(log n) time, unlike linear search which checks each element in O(n).

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