Count Occurrences of Element in Sorted Array in DSA Javascript - Execution Trace

Concept Flow - Count Occurrences of Element in Sorted Array

Start

↓

Input: sorted array, target element

↓

Find first occurrence index using binary search

↓

Find last occurrence index using binary search

↓

Calculate count = last - first + 1

↓

Return count

↓

End

We use two binary searches to find the first and last positions of the target element, then calculate how many times it appears.

Execution Sample

DSA Javascript

function countOccurrences(arr, target) {
  const first = findFirst(arr, target);
  if (first === -1) return 0;
  const last = findLast(arr, target);
  return last - first + 1;
}

This code finds the first and last positions of the target in the sorted array and returns the count of occurrences.

Execution Table

Step	Operation	Array State	Pointer Changes	Visual State
1	Start countOccurrences(arr, 2)	[1,2,2,2,3,4,5]	first = undefined, last = undefined	Array unchanged
2	Call findFirst(arr, 2)	[1,2,2,2,3,4,5]	low=0, high=6, mid=3	Check arr[3]=2 matches target
3	findFirst: arr[3]=2 matches target	Same	high=2 (move left)	Searching left half
4	findFirst: low=0, high=2, mid=1	Same	arr[1]=2 matches target	high=0 (move left)
5	findFirst: low=0, high=0, mid=0	Same	arr[0]=1 < target	low=1 (move right)
6	findFirst: low=1, high=0 (low>high)	Same	Return first=1	First occurrence at index 1
7	Call findLast(arr, 2)	[1,2,2,2,3,4,5]	low=0, high=6, mid=3	Check arr[3]=2 matches target
8	findLast: arr[3]=2 matches target	Same	low=4 (move right)	Searching right half
9	findLast: low=4, high=6, mid=5	Same	arr[5]=4 > target	high=4 (move left)
10	findLast: low=4, high=4, mid=4	Same	arr[4]=3 > target	high=3 (move left)
11	findLast: low=4, high=3 (low>high)	Same	Return last=3	Last occurrence at index 3
12	Calculate count = last - first + 1	Same	count = 3 - 1 + 1 = 3	Count of 2 is 3
13	Return count=3	Same	Function ends	Final result: 3 occurrences

💡 Binary searches end when low > high; count calculated from first and last indices.

Variable Tracker

Variable	Start	After Step 2	After Step 6	After Step 7	After Step 11	After Step 12	Final
first	undefined	undefined	1	1	1	1	1
last	undefined	undefined	undefined	undefined	3	3	3
low	undefined	0	1	0	4	4	4
high	undefined	6	0	6	3	3	3
mid	undefined	3	0	3	4	4	4
count	undefined	undefined	undefined	undefined	undefined	3	3

Key Moments - 3 Insights

Why do we need two binary searches instead of one?

One binary search finds the first occurrence (step 6), the other finds the last occurrence (step 11). This ensures we count all occurrences between these indices.

What happens if the target is not found?

If findFirst returns -1 (not found), countOccurrences returns 0 immediately (step 2), skipping findLast.

Why does findFirst move 'high' when arr[mid] equals target?

To find the first occurrence, we continue searching left (lower indices) by moving 'high' left (step 3 and 4) even if we find the target.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'first' after step 6?

A 1

B 0

C -1

D 3

💡 Hint

Check the 'Pointer Changes' column at step 6 where findFirst returns the first occurrence index.

At which step does the condition low > high become true for findLast?

A Step 9

B Step 10

C Step 11

D Step 12

💡 Hint

Look at the 'Pointer Changes' column for findLast where low=4 and high=3 indicating low > high.

If the target was not in the array, what would be the output of countOccurrences?

A Length of the array

B 0

C -1

D 1

💡 Hint

Refer to step 2 where if findFirst returns -1, countOccurrences returns 0 immediately.

Concept Snapshot

Count Occurrences in Sorted Array:
- Use binary search twice: findFirst and findLast occurrence
- findFirst moves left when target found
- findLast moves right when target found
- Count = last - first + 1
- If not found, count = 0

Full Transcript

To count how many times an element appears in a sorted array, we use two binary searches. First, find the first position where the element appears. Then, find the last position. The count is the difference plus one. If the element is not found, return zero. This method is efficient because it uses binary search, which runs fast on sorted arrays.

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