First and Last Occurrence of Element in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

public static int firstOccurrence(int[] arr, int target) {
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] == [1]) {
            return i;
        }
    }
    return -1;
}

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

Ai

Btarget

Carr

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the loop index i instead of target for comparison.

Comparing arr[i] with arr instead of target.

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

2fill in blank

medium

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

DSA Java

public static int lastOccurrence(int[] arr, int target) {
    int result = -1;
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] == [1]) {
            result = i;
        }
    }
    return result;
}

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

Ai

Barr

Ctarget

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using the loop index i instead of target for comparison.

Not updating the result variable inside the loop.

We update result whenever arr[i] equals target to track the last occurrence index.

3fill in blank

hard

Fix the error in the code to correctly find the first occurrence of target.

DSA Java

public static int firstOccurrence(int[] arr, int target) {
    int i = 0;
    while (i < arr.length) {
        if (arr[i] [1] target) {
            return i;
        }
        i++;
    }
    return -1;
}

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

A==

Btarget

C!=

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of '==' in the if condition.

Using '!=' which checks for inequality.

The if condition must use '==' to compare values, not '=' which is assignment.

4fill in blank

hard

Fill the blank to create a method that returns the first and last occurrence indices of target in the array.

DSA Java

public static int[] firstAndLastOccurrence(int[] arr, int target) {
    int first = -1;
    int last = -1;
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] [1] target) {
            if (first == -1) {
                first = i;
            }
            last = i;
        }
    }
    return new int[] {first, last};
}

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using inequality operator '!=' instead of equality.

Using comparison operators '<' or '>' which are incorrect here.

We check if arr[i] equals target to update first and last occurrence indices.

5fill in blank

hard

Fill all three blanks to create a method that returns a map with keys "first" and "last" for the first and last occurrence indices of target.

DSA Java

import java.util.Map;
import java.util.HashMap;

public static Map<String, Integer> findOccurrences(int[] arr, int target) {
    Map<String, Integer> result = new HashMap<>();
    int first = -1;
    int last = -1;
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] [1] target) {
            if (first [2] -1) {
                first = i;
            }
            last = i;
        }
    }
    result.put("[3]", first);
    result.put("last", last);
    return result;
}

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

A==

B!=

C<

Dfirst

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '!=' to check first index.

Using wrong key string instead of "first" in the map.

We check equality with '==', check if first is equal to -1 with '!=', and use "first" as the key string.

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