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

The Big Idea

What if you could count repeated items in a huge sorted list in just a few steps instead of checking every single one?

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 scanning through the entire list is slow and tiring. It's easy to lose track or make mistakes, especially if the list is very long. This wastes time and can cause errors.

The Solution

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

Before vs After

✗ Before

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

✓ After

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

What It Enables

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

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 whole 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 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