Count Occurrences of Element in Sorted Array in DSA C++ - Execution Trace

Concept Flow - Count Occurrences of Element in Sorted Array

Start

↓

Input: Sorted Array, Target Element

↓

Find First Occurrence Index

↓

Find Last Occurrence Index

↓

Calculate Count = Last - First + 1

↓

Return Count

↓

End

We find where the target first appears, then where it last appears, then count how many times it occurs.

Execution Sample

DSA C++

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

This code finds the first and last positions of x in a sorted array and returns how many times x appears.

Execution Table

Step	Operation	Index Checked	Comparison Result	Pointer Changes	Visual State
1	Find First Occurrence	mid=4	arr[4]=5 == 5	high=4	[1,2,3,4,5,5,5,6,7]
2	Find First Occurrence	mid=1	arr[1]=2 < 5	low=2	[1,2,3,4,5,5,5,6,7]
3	Find First Occurrence	mid=3	arr[3]=4 < 5	low=4	[1,2,3,4,5,5,5,6,7]
4	Find First Occurrence	mid=4	arr[4]=5 == 5	high=4	[1,2,3,4,5,5,5,6,7]
5	Find First Occurrence	mid=4	arr[4]=5 == 5	high=3	[1,2,3,4,5,5,5,6,7]
6	First Occurrence Found	index=4	-	-	[1,2,3,4,5,5,5,6,7]
7	Find Last Occurrence	mid=6	arr[6]=5 == 5	low=7	[1,2,3,4,5,5,5,6,7]
8	Find Last Occurrence	mid=7	arr[7]=6 > 5	high=6	[1,2,3,4,5,5,5,6,7]
9	Find Last Occurrence	mid=6	arr[6]=5 == 5	low=7	[1,2,3,4,5,5,5,6,7]
10	Last Occurrence Found	index=6	-	-	[1,2,3,4,5,5,5,6,7]
11	Calculate Count	-	-	-	Count = 6 - 4 + 1 = 3
12	Return Count	-	-	-	3

💡 Search ends when low > high or first/last occurrence found.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9	After Step 10	Final
low (first)	0	0	2	4	4	4	4	4	4	4	4	4
high (first)	8	4	4	4	3	3	3	3	3	3	3	3
mid (first)	-	4	1	3	4	4	-	-	-	-	-	-
firstIndex	-1	-1	-1	-1	-1	4	4	4	4	4	4	4
low (last)	0	0	0	0	0	0	0	7	7	7	7	7
high (last)	8	8	8	8	8	8	8	8	6	6	6	6
mid (last)	-	-	-	-	-	-	-	6	7	6	-	-
lastIndex	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	6	6
count	-	-	-	-	-	-	-	-	-	-	-	3

Key Moments - 3 Insights

Why do we update 'high' when arr[mid] == x in findFirst?

Because we want to find the earliest index where x appears, so we move 'high' left to continue searching earlier positions (see steps 1,4,5).

Why do we update 'low' when arr[mid] == x in findLast?

Because we want to find the last occurrence, so we move 'low' right to search later positions (see steps 7,9).

What if the element is not found at all?

The first occurrence search returns -1, so countOccurrences returns 0 immediately (not shown in this example).

Visual Quiz - 3 Questions

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

A 4

B 3

C -1

D 6

💡 Hint

Check the 'firstIndex' variable in variable_tracker after step 6.

At which step does the 'lastIndex' get assigned its final value?

A Step 9

B Step 10

C Step 7

D Step 11

💡 Hint

Look at the execution_table row where 'Last Occurrence Found' is noted.

If the target element was not in the array, what would the count be?

A 1

B -1

C 0

D Array length

💡 Hint

Refer to the key moment about element not found and the early return in code.

Concept Snapshot

Count Occurrences in Sorted Array:
- Use binary search twice: find first and last occurrence of target.
- For first occurrence: move high when arr[mid] == target.
- For last occurrence: move low when arr[mid] == target.
- Count = lastIndex - firstIndex + 1.
- If not found, count is 0.

Full Transcript

To count how many times an element appears in a sorted array, we find the first and last positions where it appears using binary search. We adjust pointers to narrow down the search: for the first occurrence, we move the high pointer left when we find the target; for the last occurrence, we move the low pointer right. The count is the difference between these positions plus one. If the element is not found, the count is zero.

Practice

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

easy

A. It reduces the search time from linear to logarithmic

B. It sorts the array automatically

C. It uses less memory than linear search

D. It works only for unsorted arrays

Solution

1. Step 1: Understand linear vs binary search time

   Linear search checks each element one by one, taking O(n) time. Binary search divides the search space in half each time, taking O(log n) time.

2. Step 2: Apply binary search to counting occurrences

   Using binary search to find the first and last occurrence of the element reduces the total search time significantly compared to scanning the whole array.

3. Final Answer:

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

4. Quick Check:

   Binary search = faster search [OK]

Hint: Binary search halves search space, speeding up counting [OK]

Common Mistakes:

• Thinking binary search sorts the array
• Confusing memory usage with speed
• Assuming binary search works on unsorted arrays

2. Which of the following C++ function signatures is correct for finding the first occurrence of an element in a sorted array using binary search?

easy

A. int firstOccurrence(int arr[], int x)

B. void firstOccurrence(int arr[], int n, int x)

C. int firstOccurrence(int arr, int n, int x)

D. int firstOccurrence(int arr[], int n, int x)

Solution

1. Step 1: Check parameter types and count

   The function needs the array (int arr[]), its size (int n), and the target element (int x).

2. Step 2: Check return type

   It should return an int index of the first occurrence, so return type is int.

3. Final Answer:

   int firstOccurrence(int arr[], int n, int x) -> Option D

4. Quick Check:

   Correct parameters and return type [OK]

Hint: Function returns index, needs array, size, and target [OK]

Common Mistakes:

• Missing array size parameter
• Wrong return type (void instead of int)
• Incorrect array parameter type

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

int arr[] = {1, 2, 2, 2, 3, 4, 5};
int n = 7;
int x = 2;

int countOccurrences(int arr[], int n, int x) {
    int first = -1, last = -1;
    int low = 0, high = n - 1;
    // Find first occurrence
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == x) {
            first = mid;
            high = mid - 1;
        } else if (arr[mid] < x) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    if (first == -1) return 0;
    low = first; high = n - 1;
    // Find last occurrence
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == x) {
            last = mid;
            low = mid + 1;
        } else if (arr[mid] < x) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return last - first + 1;
}

std::cout << countOccurrences(arr, n, x) << std::endl;

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Find first occurrence of 2

   Binary search finds first 2 at index 1.

2. Step 2: Find last occurrence of 2

   Binary search finds last 2 at index 3.

3. Step 3: Calculate count

   Count = last - first + 1 = 3 - 1 + 1 = 3.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   Occurrences of 2 = 3 [OK]

Hint: Count = last index - first index + 1 [OK]

Common Mistakes:

• Returning last index instead of count
• Not updating low and high correctly
• Returning 0 when element exists

4. Identify the bug in the following code for counting occurrences of an element in a sorted array:

int countOccurrences(int arr[], int n, int x) {
    int first = -1, last = -1;
    int low = 0, high = n - 1;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == x) {
            first = mid;
            high = mid - 1;
        } else if (arr[mid] < x) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    if (first == -1) return 0;
    low = first; high = n - 1;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == x) {
            last = mid;
            low = mid + 1;
        } else if (arr[mid] < x) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return last - first + 1;
}

medium

A. Returning last - first instead of last - first + 1

B. Using 'while (low < high)' instead of 'while (low <= high)' causes missing last element

C. Incorrect calculation of mid index

D. Not initializing 'first' and 'last' to zero

Solution

1. Step 1: Analyze loop condition

   The loops use 'while (low < high)', which misses checking when low equals high, possibly skipping the last element.

2. Step 2: Correct loop condition

   Changing to 'while (low <= high)' ensures all elements are checked, including the last one.

3. Final Answer:

   Using 'while (low < high)' instead of 'while (low <= high)' causes missing last element -> Option B

4. Quick Check:

   Loop condition must include equality [OK]

Hint: Use <= in binary search loops to include all elements [OK]

Common Mistakes:

• Using < instead of <= in loops
• Not updating first and last correctly
• Miscomputing mid index

5. Given a sorted array with possible duplicates, how would you modify the binary search approach to count occurrences of multiple different elements efficiently without repeating the search for each element separately?

hard

A. Sort the array again before each search

B. Run binary search separately for each element to find counts

C. Precompute a frequency map using a single pass, then query counts in O(1)

D. Use linear search for each element to count occurrences

Solution

1. Step 1: Understand repeated binary search cost

   Running binary search for each element separately costs O(m log n) for m queries, which can be expensive.

2. Step 2: Use frequency map for efficiency

   A single pass over the sorted array can build a frequency map (element to count) in O(n). Then queries are O(1) each.

3. Final Answer:

   Precompute a frequency map using a single pass, then query counts in O(1) -> Option C

4. Quick Check:

   Frequency map = fast multiple queries [OK]

Hint: Build frequency map once, then answer queries instantly [OK]

Common Mistakes:

• Running binary search repeatedly for each query
• Sorting array multiple times unnecessarily
• Using linear search for each query