First and Last Occurrence of Element in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find first and last occurrence indexes

What is the output of the following Java code that finds the first and last occurrence of the number 3 in the array?

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] arr = {1, 3, 5, 3, 7, 3, 9};
        int first = -1, last = -1;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == 3) {
                if (first == -1) first = i;
                last = i;
            }
        }
        System.out.println("First: " + first + ", Last: " + last);
    }
}

A First: 2, Last: 5

B First: 1, Last: 3

C First: 1, Last: 5

D First: 0, Last: 6

💡 Hint

Look for the first and last index where the value 3 appears in the array.

The number 3 appears at indexes 1, 3, and 5. The first occurrence is at index 1 and the last at index 5.

🧠 Conceptual

intermediate

Understanding first and last occurrence logic

Which statement correctly describes how to find the first and last occurrence of an element in an array?

A Traverse the array twice, first to find last occurrence, second to find first occurrence.

B Traverse the array once, update first occurrence only if not set, update last occurrence every time the element is found.

C Use a binary search twice, once for first occurrence and once for last occurrence, only if array is unsorted.

D Traverse the array once, update first and last occurrence only when the element is found at the middle index.

💡 Hint

Think about how to track first and last positions in one pass.

By traversing once, you can set the first occurrence when you find the element the first time and keep updating the last occurrence whenever you find it again.

❓ Predict Output

advanced

Output of binary search for first occurrence

What is the output of this Java code that uses binary search to find the first occurrence of 4 in a sorted array?

DSA Java

public class Main {
    public static int firstOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                high = mid - 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }
    public static void main(String[] args) {
        int[] arr = {1, 2, 4, 4, 4, 5, 6};
        System.out.println(firstOccurrence(arr, 4));
    }
}

A 2

B 4

C 3

D -1

💡 Hint

Binary search narrows down to the leftmost index where 4 appears.

The first occurrence of 4 is at index 2 in the sorted array.

❓ Predict Output

advanced

Output of binary search for last occurrence

What is the output of this Java code that uses binary search to find the last occurrence of 4 in a sorted array?

DSA Java

public class Main {
    public static int lastOccurrence(int[] arr, int x) {
        int low = 0, high = arr.length - 1, result = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == x) {
                result = mid;
                low = mid + 1;
            } else if (arr[mid] < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }
    public static void main(String[] args) {
        int[] arr = {1, 2, 4, 4, 4, 5, 6};
        System.out.println(lastOccurrence(arr, 4));
    }
}

A 4

B 3

C 2

D -1

💡 Hint

Binary search moves right to find the last occurrence index.

The last occurrence of 4 is at index 4 in the sorted array.

🧠 Conceptual

expert

Time complexity of finding first and last occurrence

What is the time complexity of finding the first and last occurrence of an element in a sorted array using binary search?

A O(log n) for each search, total O(log n)

B O(n) because you must scan the entire array

C O(n log n) because binary search is repeated multiple times

D O(log n) for each search, total O(log n * 2) which is O(log n)

💡 Hint

Consider that you do two binary searches independently.

Each binary search takes O(log n), doing two searches is 2 * O(log n) which simplifies to O(log n).

Practice

1. What does the term first occurrence of an element in an array mean?

easy

A. The index of the first time the element appears in the array

B. The total count of the element in the array

C. The last index where the element appears

D. The sum of all elements before the element

Solution

1. Step 1: Understand the meaning of first occurrence

   The first occurrence is the earliest position where the element appears in the array, starting from index 0.

2. Step 2: Compare with other options

   Counting or last index are different concepts; sum is unrelated.

3. Final Answer:

   The index of the first time the element appears in the array -> Option A

4. Quick Check:

   First occurrence = first index found [OK]

Hint: First occurrence means earliest index found [OK]

Common Mistakes:

• Confusing first occurrence with count
• Mixing first occurrence with last occurrence
• Thinking it means sum or total

2. Which of the following Java code snippets correctly finds the first occurrence of target in array arr?

easy

A. for(int i=0; i

B. for(int i=arr.length-1; i>=0; i--) { if(arr[i] == target) return i; } return -1;

C. int index = arr.indexOf(target); return index;

D. while(arr.contains(target)) { return arr.indexOf(target); }

Solution

1. Step 1: Check loop direction and condition

   for(int i=0; i

2. Step 2: Analyze other options

   for(int i=arr.length-1; i>=0; i--) { if(arr[i] == target) return i; } return -1; loops backward, so finds last occurrence. int index = arr.indexOf(target); return index; uses invalid method for arrays. while(arr.contains(target)) { return arr.indexOf(target); } has incorrect syntax and logic.

3. Final Answer:

   for(int i=0; i -> Option A

4. Quick Check:

   Loop start to end returns first match [OK]

Hint: Loop from start to find first occurrence [OK]

Common Mistakes:

• Looping backward to find first occurrence
• Using methods not available for arrays
• Incorrect loop conditions

3. What is the output of the following Java code?

int[] arr = {5, 3, 7, 3, 9};
int target = 3;
int first = -1, last = -1;
for(int i=0; i

medium

A. First: 0, Last: 4

B. First: 1, Last: 3

C. First: 3, Last: 1

D. First: -1, Last: -1

Solution

1. Step 1: Trace the loop to find first and last indices

   At i=1, arr[1]=3 matches target, first=-1 so first=1, last=1. At i=3, arr[3]=3 matches, last=3 updated.

2. Step 2: Confirm final values

   First occurrence is index 1, last occurrence is index 3.

3. Final Answer:

   First: 1, Last: 3 -> Option B

4. Quick Check:

   First=1, Last=3 from loop [OK]

Hint: Update last on every match, first only once [OK]

Common Mistakes:

• Swapping first and last values
• Not updating last occurrence correctly
• Initializing first incorrectly

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

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

medium

A. Missing return type in method

B. Variable last is not initialized

C. Loop should run backwards from end to start

D. Break statement stops loop too early, finds first not last occurrence

Solution

1. Step 1: Analyze loop and break usage

   The break inside if stops loop at first match, so last is actually first occurrence, not last.

2. Step 2: Understand correct approach

   To find last occurrence, loop should continue through entire array without break, updating last each time.

3. Final Answer:

   Break statement stops loop too early, finds first not last occurrence -> Option D

4. Quick Check:

   Break stops loop early, wrong for last occurrence [OK]

Hint: Don't break loop when searching last occurrence [OK]

Common Mistakes:

• Using break inside loop for last occurrence
• Looping forward but breaking early
• Confusing first and last occurrence logic

5. Given a sorted array arr with possible duplicates, which approach efficiently finds the first and last occurrence of target in O(log n) time?

hard

A. Use a HashMap to store indices of all elements

B. Traverse array from start to end and record first and last indices

C. Use binary search twice: once for first occurrence, once for last occurrence

D. Sort the array again and then find occurrences

Solution

1. Step 1: Understand array is sorted

   Sorted array allows binary search to find elements quickly in O(log n) time.

2. Step 2: Apply binary search twice

   One binary search finds first occurrence by moving left on matches; another finds last occurrence by moving right.

3. Step 3: Compare with other options

   Linear traversal is O(n), HashMap adds overhead, sorting again is unnecessary.

4. Final Answer:

   Use binary search twice: once for first occurrence, once for last occurrence -> Option C

5. Quick Check:

   Binary search twice = O(log n) efficient [OK]

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

Common Mistakes:

• Using linear search on sorted array
• Trying to sort array again unnecessarily
• Using extra data structures without need