First and Last Occurrence of Element in DSA Java - Execution Trace

Concept Flow - First and Last Occurrence of Element

Start: Given array and target element

↓

Initialize first = -1, last = -1

↓

Traverse array from start to end

↓

For each element: Is element == target?

↓

Update first if first == -1

↓

Update last

↓

Continue until end of array

↓

Return first and last indices

↓

Done

We scan the array once, updating first and last positions when we find the target element.

Execution Sample

DSA Java

int first = -1, last = -1;
for (int i = 0; i < arr.length; i++) {
  if (arr[i] == target) {
    if (first == -1) first = i;
    last = i;
  }
}
System.out.println(first + " " + last);

This code finds the first and last positions of the target element in the array.

Execution Table

Step	Index i	Element arr[i]	Condition arr[i] == target	First	Last	Visual State
1	0	2	False	-1	-1	[2, 3, 5, 3, 7, 3]
2	1	3	True	1	1	[2, 3, 5, 3, 7, 3]
3	2	5	False	1	1	[2, 3, 5, 3, 7, 3]
4	3	3	True	1	3	[2, 3, 5, 3, 7, 3]
5	4	7	False	1	3	[2, 3, 5, 3, 7, 3]
6	5	3	True	1	5	[2, 3, 5, 3, 7, 3]
7	End	-	-	1	5	[2, 3, 5, 3, 7, 3]

💡 Reached end of array, no more elements to check.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
i	N/A	0	1	2	3	4	5	End
first	-1	-1	1	1	1	1	1	1
last	-1	-1	1	1	3	3	5	5

Key Moments - 3 Insights

Why do we check if first == -1 before updating first?

Because first should store the earliest index where target appears. We only set it once at the first occurrence (see execution_table row 2).

Why do we update last every time we find the target?

Because last should always point to the most recent occurrence of the target (see execution_table rows 2, 4, 6).

What if the target is not in the array at all?

Both first and last remain -1, indicating target not found (would be shown if no True condition in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'first' after step 4?

A 3

B 1

C -1

D 5

💡 Hint

Check the 'First' column at step 4 in execution_table.

At which step does the 'last' variable get updated to 5?

A Step 3

B Step 4

C Step 6

D Step 2

💡 Hint

Look at the 'Last' column and find when it changes to 5 in execution_table.

If the target element was 10 (not in array), what would be the final values of first and last?

A -1 and -1

B 0 and 0

C 1 and 5

D Array length and array length

💡 Hint

Recall the initial values and that they only change when target is found.

Concept Snapshot

First and Last Occurrence of Element:
- Initialize first and last as -1
- Traverse array once
- If element == target:
  - Set first if not set
  - Update last every time
- After traversal, first and last hold positions or -1 if not found

Full Transcript

This concept finds the first and last positions of a target element in an array. We start with first and last set to -1, meaning not found. We scan the array from start to end. Each time we find the target, if first is still -1, we set it to current index. We always update last to current index when target matches. After scanning, first and last show the earliest and latest positions of the target. If target is not found, both remain -1.

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