First and Last Occurrence of Element in DSA Java - Build from Scratch

First and Last Occurrence of Element

📖 Scenario: You are working with a list of product IDs sold in a store during a day. You want to find the first and last time a specific product was sold.

🎯 Goal: Build a Java program that finds the first and last occurrence positions of a given product ID in a list of product IDs.

📋 What You'll Learn

Create an integer array called productIds with exact values

Create an integer variable called targetProduct with a specific value

Use a for loop to find the first occurrence index of targetProduct

Use a for loop to find the last occurrence index of targetProduct

Print the first and last occurrence indices exactly as specified

💡 Why This Matters

🌍 Real World

Finding the first and last occurrence of an item in a list is useful in sales data analysis, inventory tracking, and event logs.

💼 Career

This skill helps in data processing tasks, debugging, and writing efficient search algorithms in software development.

1

Create the product IDs array

Create an integer array called productIds with these exact values: 3, 5, 7, 5, 9, 5, 2

DSA Java

// Your code here

Hint

Use curly braces {} to list the values inside the array.

2

Set the target product ID

Create an integer variable called targetProduct and set it to 5

DSA Java

int[] productIds = {3, 5, 7, 5, 9, 5, 2};
// Your code here

Hint

Use int targetProduct = 5; to create the variable.

3

Find first and last occurrence of the target product

Use a for loop with variable i from 0 to productIds.length to find the first occurrence index of targetProduct and store it in firstIndex. Then use another for loop with variable j from productIds.length - 1 down to 0 to find the last occurrence index of targetProduct and store it in lastIndex. Initialize firstIndex and lastIndex to -1 before the loops.

DSA Java

int[] productIds = {3, 5, 7, 5, 9, 5, 2};
int targetProduct = 5;
// Your code here

Hint

Use one loop from start to find first occurrence, and one loop from end to find last occurrence.

4

Print the first and last occurrence indices

Print the first occurrence index using System.out.println("First occurrence: " + firstIndex); and the last occurrence index using System.out.println("Last occurrence: " + lastIndex);

DSA Java

int[] productIds = {3, 5, 7, 5, 9, 5, 2};
int targetProduct = 5;
int firstIndex = -1;
int lastIndex = -1;

for (int i = 0; i = 0; j--) {
    if (productIds[j] == targetProduct) {
        lastIndex = j;
        break;
    }
}
// Your code here

Hint

Use System.out.println to print the results exactly as shown.

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