Count Occurrences of Element in Sorted Array in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the first occurrence index of the target in the sorted array.

DSA Java

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

Drag options to blanks, or click blank then click option'

A*

B/

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' causes infinite loop or wrong index.

Not updating high pointer after finding target.

When we find the target, we move the high pointer to mid - 1 to continue searching on the left side for the first occurrence.

2fill in blank

medium

Complete the code to find the last occurrence index of the target in the sorted array.

DSA Java

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

Drag options to blanks, or click blank then click option'

A-

B/

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causes infinite loop or wrong index.

Not updating low pointer after finding target.

When the target is found, move the low pointer to mid + 1 to search the right side for the last occurrence.

3fill in blank

hard

Fix the error in the countOccurrences method to correctly calculate the count of target in the array.

DSA Java

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

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' causes wrong count.

Forgetting to add 1 after subtraction.

Count is last index minus first index plus one, so use subtraction between last and first.

4fill in blank

hard

Fill both blanks to create a map of each unique element to its count in the sorted array.

DSA Java

Map<Integer, Integer> countMap = new HashMap<>();
for (int num : arr) {
    countMap.put(num, countMap.getOrDefault(num, [1]) [2] 1);
}

Drag options to blanks, or click blank then click option'

A0

B1

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as default causes counts to be off by one.

Using '-' instead of '+' causes wrong counts.

Use 0 as default count and add 1 for each occurrence.

5fill in blank

hard

Fill both blanks to create a method that returns a map of elements to their counts using binary search methods.

DSA Java

Map<Integer, Integer> countOccurrencesMap(int[] arr) {
    Map<Integer, Integer> map = new HashMap<>();
    Set<Integer> unique = new HashSet<>();
    for (int num : arr) {
        if (!unique.contains(num)) {
            unique.add(num);
            int count = lastOccurrence(arr, num) [1] firstOccurrence(arr, num) [2] 1;
            map.put(num, count);
        }
    }
    return map;
}

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' for first operation.

Forgetting to add 1 after subtraction.

Count is last minus first plus one; the plus one is added after subtraction.

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