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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of counting occurrences using binary search

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

DSA Java

public class CountOccurrences {
    public static int countOccurrences(int[] arr, int x) {
        int first = firstOccurrence(arr, x);
        if (first == -1) return 0;
        int last = lastOccurrence(arr, x);
        return last - first + 1;
    }

    private static int firstOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                high = mid - 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    private static int lastOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                low = mid + 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 3, 3, 4, 5};
        System.out.println(countOccurrences(arr, 3));
    }
}

A 1

B 2

C 0

D 3

💡 Hint

Count how many times 3 appears consecutively in the sorted array.

The code finds the first and last positions of 3 using binary search. The first occurrence is at index 2, the last at index 4. So, count = 4 - 2 + 1 = 3.

🧠 Conceptual

intermediate

Understanding the time complexity of counting occurrences

What is the time complexity of counting occurrences of an element in a sorted array using binary search for first and last occurrences?

A O(n log n)

B O(n)

C O(log n)

D O(1)

💡 Hint

Binary search divides the search space in half each step.

Finding first and last occurrences each takes O(log n) time. So total is O(log n) + O(log n) = O(log n).

🔧 Debug

advanced

Identify the error in counting occurrences code

What error will the following code produce when counting occurrences of 7 in the array?

DSA Java

public class CountOccurrencesBug {
    public static int countOccurrences(int[] arr, int x) {
        int first = firstOccurrence(arr, x);
        int last = lastOccurrence(arr, x);
        return last - first + 1;
    }

    private static int firstOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                high = mid - 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    private static int lastOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                low = mid + 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 3, 3, 4, 5};
        System.out.println(countOccurrences(arr, 7));
    }
}

A Returns 1

B Returns -1

C Returns 0

D Throws ArrayIndexOutOfBoundsException

💡 Hint

Check what happens when the element is not found in the array.

When the element 7 is not found, firstOccurrence returns -1. The code does not check this before calculating last - first + 1, so last is also -1. The result is -1 - (-1) + 1 = 1.

❓ Predict Output

advanced

Output of counting occurrences with duplicates at edges

What is the output of the code counting occurrences of 2 in the array?

DSA Java

public class CountOccurrencesEdges {
    public static int countOccurrences(int[] arr, int x) {
        int first = firstOccurrence(arr, x);
        if (first == -1) return 0;
        int last = lastOccurrence(arr, x);
        return last - first + 1;
    }

    private static int firstOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                high = mid - 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    private static int lastOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                low = mid + 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] arr = {2, 2, 2, 3, 4, 5, 6};
        System.out.println(countOccurrences(arr, 2));
    }
}

A 3

B 2

C 1

D 0

💡 Hint

Count how many times 2 appears at the start of the array.

The element 2 appears at indices 0, 1, and 2. So count = 2 - 0 + 1 = 3.

🧠 Conceptual

expert

Why binary search is preferred over linear scan for counting occurrences in sorted arrays

Why is binary search a better approach than a linear scan to count occurrences of an element in a sorted array?

A Binary search uses less memory than linear scan.

B Binary search reduces time complexity from O(n) to O(log n), making it faster for large arrays.

C Linear scan cannot find the element if it appears multiple times.

D Binary search works only on unsorted arrays, making it more versatile.

💡 Hint

Think about how many elements each method checks in the worst case.

Binary search divides the search space in half each step, reducing the number of checks drastically compared to linear scan which checks every element.

Practice

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

easy

A. It uses less memory than linear search

B. It reduces the search time to logarithmic complexity

C. It sorts the array automatically

D. It works only for unsorted arrays

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 search

   Linear search checks each element one by one, which takes longer for large arrays.

3. Final Answer:

   It reduces the search time to logarithmic complexity -> Option B

4. Quick Check:

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

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

Common Mistakes:

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

2. Which Java method signature correctly represents a function to count occurrences of an integer target in a sorted array arr?

easy

A. public static int countOccurrences(int[] arr, int target)

B. public int countOccurrences(int target, int[] arr)

C. static void countOccurrences(int[] arr, int target)

D. int countOccurrences(int[] arr, int target)

Solution

1. Step 1: Check method return type and parameters

   The method should return an integer count and accept an integer array and target integer as parameters.

2. Step 2: Verify static and access modifiers

   Static is common for utility methods; public allows access. The order of parameters matters for clarity.

3. Final Answer:

   public static int countOccurrences(int[] arr, int target) -> Option A

4. Quick Check:

   Correct signature returns int and takes (int[], int) [OK]

Hint: Return count as int, parameters: array then target [OK]

Common Mistakes:

• Wrong parameter order
• Void return type instead of int
• Missing static keyword when needed

3. What is the output of the following Java code snippet?

int[] arr = {1, 2, 2, 2, 3, 4, 5};
int target = 2;
int count = countOccurrences(arr, target);
System.out.println(count);

public static int countOccurrences(int[] arr, int target) {
    int first = firstOccurrence(arr, target);
    if (first == -1) return 0;
    int last = lastOccurrence(arr, target);
    return last - first + 1;
}

public static int firstOccurrence(int[] arr, int target) {
    int low = 0, high = arr.length - 1, 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;
}

public static int lastOccurrence(int[] arr, int target) {
    int low = 0, high = arr.length - 1, 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;
}

medium

A. 2

B. 4

C. 3

D. 0

Solution

1. Step 1: Find first occurrence of 2 in array

   First occurrence is at index 1 (arr[1] = 2).

2. Step 2: Find last occurrence of 2 in array

   Last occurrence is at index 3 (arr[3] = 2).

3. Step 3: Calculate count

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

4. Final Answer:

   3 -> Option C

5. Quick Check:

   Occurrences of 2 = 3 [OK]

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

Common Mistakes:

• Confusing first and last indices
• Off-by-one errors in count calculation
• Returning 0 when element exists

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

public static int countOccurrences(int[] arr, int target) {
    int low = 0, high = arr.length - 1;
    int count = 0;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == target) {
            count++;
            low = mid + 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return count;
}

medium

A. It counts only occurrences after the first found index

B. It counts only one occurrence instead of all

C. It counts duplicates multiple times incorrectly

D. It never terminates due to infinite loop

Solution

1. Step 1: Analyze counting logic in binary search

   The code increments count only when arr[mid] == target and moves low to mid + 1, skipping elements before mid.

2. Step 2: Understand missed occurrences

   This approach counts occurrences only from the first found index forward, missing any before mid.

3. Final Answer:

   It counts only occurrences after the first found index -> Option A

4. Quick Check:

   Counting misses earlier duplicates [OK]

Hint: Counting inside binary search loop misses earlier matches [OK]

Common Mistakes:

• Assuming count increments all occurrences
• Not updating low/high correctly
• Expecting binary search to count all duplicates directly

5. Given a sorted array arr that may contain multiple occurrences of target, how can you efficiently count the occurrences if the array is very large and you want to minimize the number of comparisons?

hard

A. Use a hash map to store counts of all elements

B. Use a linear scan from start to end counting matches

C. Sort the array first, then count occurrences

D. Use two binary searches to find first and last occurrence, then subtract indices

Solution

1. Step 1: Understand the problem constraints

   Array is sorted and large, so linear scan is costly and sorting again is unnecessary.

2. Step 2: Apply binary search twice

   Use binary search to find the first occurrence index and last occurrence index of target efficiently.

3. Step 3: Calculate count

   Count = last occurrence index - first occurrence index + 1, minimizing comparisons.

4. Final Answer:

   Use two binary searches to find first and last occurrence, then subtract indices -> Option D

5. Quick Check:

   Two binary searches = efficient counting [OK]

Hint: Find first and last index with binary search [OK]

Common Mistakes:

• Using linear scan on large arrays
• Sorting already sorted array again
• Using extra memory unnecessarily