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

Count Occurrences of Element in Sorted Array

📖 Scenario: You have a sorted list of product IDs from a store's sales records. You want to find out how many times a specific product was sold.

🎯 Goal: Build a program that counts how many times a given product ID appears in the sorted array.

📋 What You'll Learn

Create an array called productIDs with the exact values: [101, 101, 102, 103, 103, 103, 104, 105]

Create a variable called targetID and set it to 103

Use a for loop with variable id to iterate over productIDs

Inside the loop, count how many times targetID appears

Print the count with console.log

💡 Why This Matters

🌍 Real World

Counting how many times a product was sold helps stores understand popular items and manage inventory.

💼 Career

This skill is useful for data analysis, software development, and any job that involves processing lists of data efficiently.

1

Create the sorted array of product IDs

Create an array called productIDs with these exact values: [101, 101, 102, 103, 103, 103, 104, 105]

DSA Javascript

// Your code here

Hint

Use const to create the array named productIDs with the given numbers.

2

Set the target product ID to count

Create a variable called targetID and set it to 103

DSA Javascript

const productIDs = [101, 101, 102, 103, 103, 103, 104, 105];
// Your code here

Hint

Use const to create targetID and assign it the value 103.

3

Count how many times the target ID appears

Create a variable called count and set it to 0. Then use a for loop with variable id to iterate over productIDs. Inside the loop, increase count by 1 each time id equals targetID.

DSA Javascript

const productIDs = [101, 101, 102, 103, 103, 103, 104, 105];
const targetID = 103;
// Your code here

Hint

Start count at 0. Use a for...of loop to check each id in productIDs. Increase count when id matches targetID.

4

Print the count of the target product ID

Use console.log to print the value of count

DSA Javascript

const productIDs = [101, 101, 102, 103, 103, 103, 104, 105];
const targetID = 103;
let count = 0;
for (const id of productIDs) {
    if (id === targetID) {
        count++;
    }
}
// Your code here

Hint

Use console.log(count); to show the number of times targetID appears.

Practice

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

easy

A. It sorts the array automatically.

B. It reduces the search time from linear to logarithmic.

C. It counts all elements without searching.

D. It works only on 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 linear vs binary search

   Linear search checks each element one by one (O(n)), binary search works in O(log n) time.

3. Final Answer:

   It reduces the search time from linear to logarithmic. -> Option B

4. Quick Check:

   Binary search = faster search [OK]

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

Common Mistakes:

• Thinking binary search sorts the array
• Assuming binary search works on unsorted arrays
• Confusing counting with searching

2. Which of the following is the correct syntax to find the middle index in a binary search on array arr between indices low and high in JavaScript?

easy

A. const mid = low + high;

B. const mid = (low + high) / 2;

C. const mid = Math.ceil((low + high) / 2);

D. const mid = Math.floor((low + high) / 2);

Solution

1. Step 1: Calculate middle index correctly

   We use Math.floor to get the integer index because array indices must be whole numbers.

2. Step 2: Avoid floating point and wrong calculations

   Using just division or Math.ceil can cause errors or out-of-bound indices.

3. Final Answer:

   const mid = Math.floor((low + high) / 2); -> Option D

4. Quick Check:

   Use Math.floor for middle index [OK]

Hint: Use Math.floor to get integer middle index [OK]

Common Mistakes:

• Not using Math.floor causing float index
• Using Math.ceil which may skip elements
• Adding indices without division

3. What will be the output of the following JavaScript code?

function countOccurrences(arr, x) {
  let low = 0, high = arr.length - 1;
  let count = 0;
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    if (arr[mid] === x) {
      count++;
      let left = mid - 1;
      while (left >= 0 && arr[left] === x) {
        count++;
        left--;
      }
      let right = mid + 1;
      while (right < arr.length && arr[right] === x) {
        count++;
        right++;
      }
      break;
    } else if (arr[mid] < x) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return count;
}

console.log(countOccurrences([1,2,2,2,3,4,5], 2));

medium

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Find one occurrence of 2 using binary search

   The binary search finds 2 at index 2 (middle of array).

2. Step 2: Count all adjacent 2s on left and right

   Left side: indices 1 and 0 checked; only index 1 has 2. Right side: index 3 has 2. Total count = 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Count adjacent 2s = 3 [OK]

Hint: Count neighbors after binary search hit [OK]

Common Mistakes:

• Counting only the found element once
• Not checking neighbors on both sides
• Incorrect loop conditions causing infinite loops

4. The following code is intended to count occurrences of x in a sorted array arr. What is the bug?

function countOccurrences(arr, x) {
  let low = 0, high = arr.length - 1;
  let count = 0;
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    if (arr[mid] === x) {
      count++;
      let left = mid - 1;
      while (arr[left] === x) {
        count++;
        left--;
      }
      let right = mid + 1;
      while (arr[right] === x) {
        count++;
        right++;
      }
      break;
    } else if (arr[mid] < x) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return count;
}

medium

A. The while loops do not check array bounds before accessing elements.

B. The binary search condition is incorrect.

C. The count variable is not initialized.

D. The function does not return any value.

Solution

1. Step 1: Analyze while loops for boundary checks

   The loops check arr[left] === x and arr[right] === x without verifying if left or right are valid indices.

2. Step 2: Identify risk of out-of-bounds access

   If left becomes -1 or right exceeds array length, accessing arr[left] or arr[right] causes errors.

3. Final Answer:

   The while loops do not check array bounds before accessing elements. -> Option A

4. Quick Check:

   Always check array bounds before access [OK]

Hint: Check array bounds before accessing elements [OK]

Common Mistakes:

• Ignoring array index limits in loops
• Assuming array access is always safe
• Not breaking loop on invalid index

5. Given a sorted array arr = [1,2,2,2,3,3,4,5,5,5,5,6], which approach is the most efficient to count occurrences of 5 without scanning the entire array?

hard

A. Use binary search to find first and last positions of 5, then calculate count.

B. Use a for loop to count all 5s by scanning the whole array.

C. Sort the array again and count 5s.

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

Solution

1. Step 1: Understand the problem constraints

   The array is sorted, so all 5s are grouped together.

2. Step 2: Use binary search to find boundaries

   Find the first index of 5 and last index of 5 using binary search, then count = last - first + 1.

3. Step 3: Compare efficiency

   This method is O(log n), faster than scanning entire array (O(n)) or sorting again (O(n log n)).

4. Final Answer:

   Use binary search to find first and last positions of 5, then calculate count. -> Option A

5. Quick Check:

   Binary search boundaries = efficient count [OK]

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

Common Mistakes:

• Scanning entire array unnecessarily
• Sorting already sorted array again
• Using extra space with hash map when not needed