Count Occurrences of Element in Sorted Array in DSA Java - Time & Space Complexity

Time Complexity: Count Occurrences of Element in Sorted Array

O(log n)

Understanding Time Complexity

We want to know how long it takes to count how many times a number appears in a sorted list.

How does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

private int findFirst(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;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

private int findLast(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;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

This code finds the first and last positions of the target number using binary search, then calculates how many times it appears.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Binary search loops in findFirst and findLast.
• How many times: Each binary search loop runs about log base 2 of n times, where n is the array size.

How Execution Grows With Input

As the list gets bigger, the number of steps grows slowly because each step cuts the search space in half.

Input Size (n)	Approx. Operations
10	About 6 steps (2 binary searches x ~3 steps each)
100	About 14 steps (2 x ~7 steps)
1000	About 20 steps (2 x ~10 steps)

Pattern observation: Doubling the input size adds only a few more steps, showing slow growth.

Final Time Complexity

Time Complexity: O(log n)

This means the time to count occurrences grows slowly as the list gets bigger, making it efficient for large lists.

Common Mistake

[X] Wrong: "Counting occurrences requires checking every element one by one, so it must be slow (O(n))."

[OK] Correct: Because the list is sorted, we can use binary search to jump quickly to the first and last positions, avoiding checking every element.

Interview Connect

Knowing how to use binary search to count occurrences shows you understand how to use sorting to speed up searching, a key skill in many coding challenges.

Self-Check

"What if the array was not sorted? How would the time complexity change?"

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