First and Last Occurrence of Element in DSA Go

Mental Model

Find where a number first appears and last appears in a list by checking each item one by one.

Analogy: Imagine looking through a row of books to find the first and last time a certain title shows up on the shelf.

Index: 0   1   2   3   4   5   6
List:  2 -> 3 -> 5 -> 3 -> 7 -> 3 -> 9
               ↑           ↑
           first?       last?

Dry Run Walkthrough

Input: list: [2, 3, 5, 3, 7, 3, 9], find first and last occurrence of 3

Goal: Find the first and last positions where 3 appears in the list

Step 1: Check element at index 0 (2), not 3

2 -> 3 -> 5 -> 3 -> 7 -> 3 -> 9
first: -1, last: -1

Why: We start from the beginning to find the first occurrence

Step 2: Check element at index 1 (3), found first occurrence

2 -> [3] -> 5 -> 3 -> 7 -> 3 -> 9
first: 1, last: -1

Why: First time we see 3, record index 1 as first occurrence

Step 3: Check element at index 2 (5), not 3

2 -> 3 -> 5 -> 3 -> 7 -> 3 -> 9
first: 1, last: -1

Why: Keep searching for last occurrence

Step 4: Check element at index 3 (3), update last occurrence

2 -> 3 -> 5 -> [3] -> 7 -> 3 -> 9
first: 1, last: 3

Why: Found 3 again, update last occurrence to index 3

Step 5: Check element at index 4 (7), not 3

2 -> 3 -> 5 -> 3 -> 7 -> 3 -> 9
first: 1, last: 3

Why: Continue searching for last occurrence

Step 6: Check element at index 5 (3), update last occurrence

2 -> 3 -> 5 -> 3 -> 7 -> [3] -> 9
first: 1, last: 5

Why: Found 3 again, update last occurrence to index 5

Step 7: Check element at index 6 (9), not 3

2 -> 3 -> 5 -> 3 -> 7 -> 3 -> 9
first: 1, last: 5

Why: Reached end of list, stop searching

Result:

2 -> 3 -> 5 -> 3 -> 7 -> 3 -> 9
first occurrence of 3 at index 1
last occurrence of 3 at index 5

Annotated Code

DSA Go

package main

import "fmt"

func firstAndLastOccurrence(arr []int, target int) (int, int) {
    first := -1
    last := -1
    for i, val := range arr {
        if val == target {
            if first == -1 {
                first = i // record first occurrence
            }
            last = i // update last occurrence
        }
    }
    return first, last
}

func main() {
    arr := []int{2, 3, 5, 3, 7, 3, 9}
    target := 3
    first, last := firstAndLastOccurrence(arr, target)
    fmt.Printf("List: %v\n", arr)
    fmt.Printf("First occurrence of %d at index %d\n", target, first)
    fmt.Printf("Last occurrence of %d at index %d\n", target, last)
}

for i, val := range arr {

iterate over each element to check for target

if val == target {

check if current element matches target

if first == -1 { first = i }

record first occurrence only once

last = i

update last occurrence whenever target found

Output

List: [2 3 5 3 7 3 9]

First occurrence of 3 at index 1

Last occurrence of 3 at index 5

Complexity Analysis

Time: O(n) because we scan the entire list once to find occurrences

Space: O(1) because we use only a few variables regardless of input size

vs Alternative: Better than searching twice separately for first and last because it uses one pass instead of two

Edge Cases

empty list

returns -1 for both first and last since target not found

DSA Go

for i, val := range arr {

target not in list

returns -1 for both first and last indicating no occurrence

DSA Go

if val == target {

target appears once

first and last both equal the single index where target appears

DSA Go

if first == -1 { first = i }

When to Use This Pattern

When asked to find first and last positions of a value in a list, use a single pass scan updating first and last indices to solve efficiently.

Common Mistakes

Mistake: Searching for first and last occurrences in separate loops doubling time
Fix: Combine both searches in one loop by updating first once and last every time target is found

Mistake: Not initializing first to -1 causing wrong first occurrence
Fix: Initialize first to -1 and only set it when target is found first time

Summary

Finds the first and last positions where a target number appears in a list.

Use when you need to know the range of positions for a value in a list quickly.

Remember to scan once, record first occurrence once, and update last occurrence every time.

Practice

1. What does the "First and Last Occurrence of Element" problem ask you to find in an array?

easy

A. The sum of all elements in the array

B. The first and last positions where a given element appears

C. The maximum and minimum values in the array

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 return the indices of the first and last occurrence of the element, not sums or counts.

3. Final Answer:

   The first and last positions where a given element appears -> Option B

4. Quick Check:

   First and last occurrence = positions [OK]

Hint: Look for positions of element, not sums or counts [OK]

Common Mistakes:

• Confusing occurrence with sum or count
• Looking for max/min instead of positions
• Ignoring the last occurrence

2. Which of the following Go code snippets correctly declares a function to find the first occurrence of an element in a slice?

easy

A. func firstOccurrence(arr []int, x int) int { for _, v := range arr { if v == x { return i } } return -1 }

B. func firstOccurrence(arr []int, x int) int { for i := range arr { if arr[i] = x { return i } } return -1 }

C. func firstOccurrence(arr []int, x int) { for i := 0; i < len(arr); i++ { if arr[i] == x { return i } } return -1 }

D. func firstOccurrence(arr []int, x int) int { for i := 0; i < len(arr); i++ { if arr[i] == x { return i } } return -1 }

Solution

1. Step 1: Check function signature and return type

   The function must return an int index, so signature must be func firstOccurrence(arr []int, x int) int.

2. Step 2: Verify loop and condition syntax

   func firstOccurrence(arr []int, x int) int { for i := 0; i < len(arr); i++ { if arr[i] == x { return i } } return -1 } uses a for loop with index and correct equality check (arr[i] == x) and returns index or -1.

3. Final Answer:

   func firstOccurrence(arr []int, x int) int { for i := 0; i < len(arr); i++ { if arr[i] == x { return i } } return -1 } -> Option D

4. Quick Check:

   Correct syntax and return type = func firstOccurrence(arr []int, x int) int { for i := 0; i < len(arr); i++ { if arr[i] == x { return i } } return -1 } [OK]

Hint: Check function return type and equality operator carefully [OK]

Common Mistakes:

• Missing return type in function signature
• Using assignment '=' instead of equality '=='
• Returning inside a void function

3. What will be the output of the following Go code snippet?

arr := []int{2, 4, 4, 4, 6, 7}
first := -1
last := -1
x := 4
for i, v := range arr {
  if v == x {
    if first == -1 {
      first = i
    }
    last = i
  }
}
fmt.Println(first, last)

medium

A. 1 3

B. 2 4

C. 1 4

D. 0 3

Solution

1. Step 1: Trace the loop to find first and last occurrence of 4

   Indices with value 4 are 1, 2, 3. First occurrence is 1, last occurrence is 3.

2. Step 2: Confirm printed values

   Variables first and last hold 1 and 3 respectively, so output is "1 3".

3. Final Answer:

   1 3 -> Option A

4. Quick Check:

   First=1, Last=3 for element 4 [OK]

Hint: Track indices where element matches, update first once, last every time [OK]

Common Mistakes:

• Confusing index positions starting at 0
• Updating first occurrence multiple times
• Not updating last occurrence correctly

4. Identify the bug in this Go function that finds the last occurrence of an element in a slice:

func lastOccurrence(arr []int, x int) int {
  last := -1
  for i := 0; i <= len(arr); i++ {
    if arr[i] == x {
      last = i
    }
  }
  return last
}

medium

A. Loop condition should be i < len(arr), not i <= len(arr) to avoid out-of-range error

B. Variable 'last' should be initialized to 0 instead of -1

C. The function should return first occurrence, not last

D. The if condition should use '!=' instead of '=='

Solution

1. Step 1: Check loop boundary conditions

   The loop uses i <= len(arr), which causes index out-of-range error because valid indices are 0 to len(arr)-1.

2. Step 2: Correct loop condition

   Change to i < len(arr) to safely access elements without error.

3. Final Answer:

   Loop condition should be i < len(arr), not i <= len(arr) to avoid out-of-range error -> Option A

4. Quick Check:

   Loop index must be less than length [OK]

Hint: Use i < len(arr) for loops to avoid index errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing last to 0 causing wrong default
• Confusing last occurrence with first

5. Given a sorted slice of integers, which approach is the most efficient to find the first and last occurrence of a target element x?

hard

A. Sort the slice again and then find occurrences linearly

B. Scan the slice from start to end once, record first and last occurrence

C. Use two binary searches: one to find first occurrence, one for last occurrence

D. Use a hash map to store all indices of elements and look up x

Solution

1. Step 1: Understand the input is sorted

   Since the slice is sorted, binary search can be used to find positions efficiently.

2. Step 2: Use two binary searches

   One binary search finds the first occurrence, another finds the last occurrence of x in O(log n) time.

3. Step 3: Compare with other options

   Linear scan is O(n), sorting again is unnecessary and costly, hash map adds extra space and complexity.

4. Final Answer:

   Use two binary searches: one to find first occurrence, one for last occurrence -> Option C

5. Quick Check:

   Sorted array + binary search = efficient [OK]

Hint: Use binary search twice on sorted data for first and last positions [OK]

Common Mistakes:

• Using linear scan on sorted data ignoring binary search
• Sorting already sorted data again
• Using extra space unnecessarily