First and Last Occurrence of Element in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the first occurrence of 'target' in the array.

DSA C++

int firstOccurrence(vector<int>& arr, int target) {
    for (int i = 0; i < arr.size(); i++) {
        if (arr[i] == [1]) {
            return i;
        }
    }
    return -1;
}

Drag options to blanks, or click blank then click option'

Aarr[i]

Btarget

Ci

Darr.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr.size() instead of target in the if condition.

Comparing i with target instead of arr[i].

We compare each element arr[i] with the target to find the first occurrence.

2fill in blank

medium

Complete the code to find the last occurrence of 'target' in the array.

DSA C++

int lastOccurrence(vector<int>& arr, int target) {
    for (int i = arr.size() - 1; i >= 0; i--) {
        if (arr[i] == [1]) {
            return i;
        }
    }
    return -1;
}

Drag options to blanks, or click blank then click option'

Aarr.size()

Bi

Ctarget

Darr[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of target in the if condition.

Comparing arr.size() with arr[i].

We compare each element arr[i] with the target starting from the end to find the last occurrence.

3fill in blank

hard

Fix the error in the binary search to find the first occurrence of 'target'.

DSA C++

int firstOccurrenceBinarySearch(vector<int>& arr, int target) {
    int low = 0, high = arr.size() - 1, result = -1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == [1]) {
            result = mid;
            high = mid - 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

Drag options to blanks, or click blank then click option'

Atarget

Bmid

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr[mid] with mid instead of target.

Updating wrong pointers after comparison.

We compare arr[mid] with the target to find the first occurrence using binary search.

4fill in blank

hard

Fill both blanks to complete the binary search for the last occurrence of 'target'.

DSA C++

int lastOccurrenceBinarySearch(vector<int>& arr, int target) {
    int low = 0, high = arr.size() - 1, result = -1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == [1]) {
            result = mid;
            low = mid [2] 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

Drag options to blanks, or click blank then click option'

Atarget

Bhigh

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' when updating low pointer.

Comparing arr[mid] with high instead of target.

When arr[mid] equals target, update result and move low pointer to mid + 1 to find last occurrence.

5fill in blank

hard

Fill all three blanks to create a function that returns both first and last occurrences of 'target'.

DSA C++

pair<int, int> findFirstAndLast(vector<int>& arr, int target) {
    int first = -1, last = -1;
    int low = 0, high = arr.size() - 1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == [1]) {
            first = mid;
            high = mid [2] 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    low = 0; high = arr.size() - 1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == [3]) {
            last = mid;
            low = mid + 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return {first, last};
}

Drag options to blanks, or click blank then click option'

Atarget

B-

C+

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' when updating high pointer in first search.

Mixing up target and mid in comparisons.

Use binary search twice: first to find first occurrence (move high to mid - 1), then last occurrence (move low to mid + 1).

Practice

1. What does the "First and Last Occurrence of Element" problem ask you to find in an array?

easy

A. The maximum and minimum elements in the array

B. The sum of all elements in the array

C. The first and last positions where a given element appears

D. The number of unique elements in the array

Solution

1. Step 1: Understand the problem statement

   The problem asks to find the first and last positions of a specific element in an array.

2. Step 2: Identify what is required

   We need to locate the earliest and latest index where the element appears.

3. Final Answer:

   The first and last positions where a given element appears -> Option C

4. Quick Check:

   First and last occurrence = positions of element [OK]

Hint: Look for earliest and latest index of element [OK]

Common Mistakes:

• Confusing with sum or max/min of array
• Returning count instead of positions
• Ignoring if element is not present

2. Which of the following C++ code snippets correctly finds the first occurrence of element x in vector arr?

easy

A. int first = 0; while(arr[first] != x) { first++; }

B. int first = -1; for(int i=0; i

C. int first = arr.size(); for(int i=0; i

D. int first = -1; for(int i=arr.size()-1; i>=0; i--) { if(arr[i] == x) { first = i; break; } }

Solution

1. Step 1: Check each option for correctness

   int first = -1; for(int i=0; i

2. Step 2: Identify errors in other options

   int first = 0; while(arr[first] != x) { first++; } may cause out-of-bound error if x not found; int first = arr.size(); for(int i=0; i=0; i--) { if(arr[i] == x) { first = i; break; } } finds last occurrence due to reverse loop.

3. Final Answer:

   int first = -1; for(int i=0; i -> Option B

4. Quick Check:

   Loop from start, break on first match = first occurrence [OK]

Hint: Loop forward, break on first match for first occurrence [OK]

Common Mistakes:

• Looping backward for first occurrence
• Not handling element not found case
• Using wrong initial index values

3. What will be the output of the following C++ code?

std::vector<int> arr = {2, 3, 5, 3, 7};
int x = 3;
int first = -1, last = -1;
for(int i=0; i<arr.size(); i++) {
  if(arr[i] == x) {
    if(first == -1) first = i;
    last = i;
  }
}
std::cout << first << " " << last << std::endl;

medium

A. 2 3

B. 0 4

C. 1 1

D. 1 3

Solution

1. Step 1: Trace the loop to find first and last occurrence of 3

   Indices: 0->2, 1->3, 2->5, 3->3, 4->7. First 3 at index 1, last 3 at index 3.

2. Step 2: Confirm output printed

   Prints first and last indices: 1 3.

3. Final Answer:

   1 3 -> Option D

4. Quick Check:

   First=1, Last=3 for element 3 [OK]

Hint: Track first set once, update last on every match [OK]

Common Mistakes:

• Confusing indices with values
• Not updating last occurrence correctly
• Printing values instead of indices

4. Identify the bug in this code snippet that tries to find the last occurrence of x in arr:

int last = -1;
for(int i=0; i<arr.size(); i++) {
  if(arr[i] = x) {
    last = i;
  }
}

medium

A. Using assignment (=) instead of comparison (==) in if condition

B. Loop should run backwards from end to start

C. Variable last should be initialized to 0

D. Missing break statement after finding last occurrence

Solution

1. Step 1: Check the if condition

   The condition uses single '=' which assigns x to arr[i], not compares.

2. Step 2: Understand impact of assignment in condition

   Assignment returns value of x, which may cause wrong behavior or infinite loop.

3. Final Answer:

   Using assignment (=) instead of comparison (==) in if condition -> Option A

4. Quick Check:

   Use '==' for comparison, not '=' [OK]

Hint: Use '==' for comparison, '=' is assignment [OK]

Common Mistakes:

• Confusing '=' and '==' in conditions
• Thinking loop direction causes bug here
• Assuming break needed for last occurrence

5. Given a sorted vector arr and an element x, which approach is the most efficient to find the first and last occurrence of x?

hard

A. Use binary search twice to find first and last positions

B. Use linear search from start to end

C. Sort the array again and then find occurrences

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

Solution

1. Step 1: Understand the array is sorted

   Sorted array allows binary search to find elements efficiently.

2. Step 2: Use binary search twice

   One binary search finds first occurrence, another finds last occurrence in O(log n) time.

3. Step 3: Compare other options

   Linear search is O(n), sorting again is unnecessary, hash map adds extra space and complexity.

4. Final Answer:

   Use binary search twice to find first and last positions -> Option A

5. Quick Check:

   Sorted array + binary search = efficient first/last occurrence [OK]

Hint: Binary search twice on sorted array for first and last occurrence [OK]

Common Mistakes:

• Using linear search on sorted array
• Sorting array again unnecessarily
• Ignoring binary search advantages