First and Last Occurrence of Element in DSA C++ - Time & Space Complexity

Time Complexity: First and Last Occurrence of Element

O(log n)

Understanding Time Complexity

We want to know how long it takes to find the first and last positions of a number in a list.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code finds the first and last positions of a target number in a sorted list using binary search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that halves the search range each time.
• How many times: The loop runs about log base 2 of n times, where n is the list size.

How Execution Grows With Input

Each time the list size doubles, the number of steps increases by only one.

Input Size (n)	Approx. Operations
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: The steps grow slowly as the list gets bigger, increasing by one each time the size doubles.

Final Time Complexity

Time Complexity: O(log n)

This means the search gets only a little slower even if the list grows a lot.

Common Mistake

[X] Wrong: "Finding first and last occurrence takes as long as checking every item one by one."

[OK] Correct: Because the list is sorted, we can jump around and cut the search area in half each time, so we don't check every item.

Interview Connect

Knowing how to quickly find positions in sorted data shows you understand efficient searching, a key skill in many coding tasks.

Self-Check

"What if the list was not sorted? How would the time complexity change?"

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