Count Occurrences of Element in Sorted Array in DSA Java - Build from Scratch

Count Occurrences of Element in Sorted Array

📖 Scenario: You have a sorted list of numbers representing daily temperatures recorded over a month. You want to find out how many days had a specific temperature.

🎯 Goal: Build a Java program that counts how many times a given temperature appears in the sorted array of temperatures.

📋 What You'll Learn

Create an integer array called temperatures with the exact values: 15, 15, 16, 16, 16, 17, 18, 18, 19, 20

Create an integer variable called targetTemp and set it to 16

Write a for loop using the variable temp to iterate over temperatures and count how many times targetTemp appears

Print the count in the format: System.out.println("Count of " + targetTemp + ": " + count);

💡 Why This Matters

🌍 Real World

Counting occurrences of specific values in sorted data is common in analyzing sensor readings, sales data, or user activity logs.

💼 Career

This skill helps in data analysis roles, software testing, and any programming job that requires processing and summarizing sorted datasets efficiently.

1

Create the sorted array of temperatures

Create an integer array called temperatures with these exact values: 15, 15, 16, 16, 16, 17, 18, 18, 19, 20

DSA Java

// Create the integer array temperatures with the given values
// Your code here

Hint

Use int[] temperatures = { ... }; to create the array with the exact numbers.

2

Set the target temperature to count

Create an integer variable called targetTemp and set it to 16

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] temperatures = {15, 15, 16, 16, 16, 17, 18, 18, 19, 20};
        // Create targetTemp variable and set it to 16
        // Your code here
    }
}

Hint

Use int targetTemp = 16; to create the variable.

3

Count how many times the target temperature appears

Write a for loop using the variable temp to iterate over temperatures and count how many times targetTemp appears. Use an integer variable called count initialized to 0 before the loop.

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] temperatures = {15, 15, 16, 16, 16, 17, 18, 18, 19, 20};
        int targetTemp = 16;
        // Initialize count to 0
        // Use for loop with variable temp to count occurrences of targetTemp
        // Your code here
    }
}

Hint

Use a for-each loop: for (int temp : temperatures) and inside it, check if temp == targetTemp to increase count.

4

Print the count of the target temperature

Print the count using System.out.println("Count of " + targetTemp + ": " + count);

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] temperatures = {15, 15, 16, 16, 16, 17, 18, 18, 19, 20};
        int targetTemp = 16;
        int count = 0;
        for (int temp : temperatures) {
            if (temp == targetTemp) {
                count++;
            }
        }
        // Print the count of targetTemp
        // Your code here
    }
}

Hint

Use System.out.println("Count of " + targetTemp + ": " + count); to print the result.

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