Array Traversal Patterns in DSA Go - Build from Scratch

Array Traversal Patterns

📖 Scenario: You are helping a small grocery store manage their daily sales data. The store records the number of items sold each day in a week.

🎯 Goal: Build a simple Go program that stores the daily sales in an array, sets a target sales number, counts how many days met or exceeded the target, and then prints the count.

📋 What You'll Learn

Create an array called dailySales with exactly 7 integers representing sales for each day: 12, 15, 7, 20, 18, 5, 10

Create an integer variable called target and set it to 10

Use a for loop with index i to traverse dailySales and count how many days have sales greater than or equal to target

Print the count of days meeting or exceeding the target using fmt.Println

💡 Why This Matters

🌍 Real World

Counting how many days meet a sales target helps store managers understand performance and plan inventory.

💼 Career

Array traversal and conditional counting are fundamental skills for data processing and analysis in software development.

1

Create the daily sales array

Create an array called dailySales with these exact 7 integers: 12, 15, 7, 20, 18, 5, 10

DSA Go

package main

import "fmt"

func main() {
    // Create the dailySales array with 7 integers
    // Your code here
}

Hint

Use Go array syntax: varName := [7]int{values}

2

Set the sales target

Create an integer variable called target and set it to 10

DSA Go

package main

import "fmt"

func main() {
    dailySales := [7]int{12, 15, 7, 20, 18, 5, 10}
    // Create the target variable and set it to 10
    // Your code here
}

Hint

Use simple variable assignment: target := 10

3

Count days meeting or exceeding the target

Use a for loop with index i to traverse dailySales and count how many days have sales greater than or equal to target. Store the count in a variable called count

DSA Go

package main

import "fmt"

func main() {
    dailySales := [7]int{12, 15, 7, 20, 18, 5, 10}
    target := 10
    // Initialize count variable
    // Use a for loop with index i to check each day's sales
    // Increase count if sales >= target
    // Your code here
}

Hint

Use for i := 0; i < len(dailySales); i++ to loop through the array

4

Print the count of days meeting or exceeding the target

Print the variable count using fmt.Println

DSA Go

package main

import "fmt"

func main() {
    dailySales := [7]int{12, 15, 7, 20, 18, 5, 10}
    target := 10
    count := 0
    for i := 0; i = target {
            count++
        }
    }
    // Print the count variable
    // Your code here
}

Hint

Use fmt.Println(count) to print the result

Practice

1. What does the following Go code snippet do?
for i := 0; i < len(arr); i++ { fmt.Println(arr[i]) }

easy

A. Prints each element of the array from start to end

B. Prints the array elements in reverse order

C. Prints only the first element of the array

D. Causes a runtime error due to index out of range

Solution

1. Step 1: Understand the loop structure

   The loop starts at index 0 and runs while i is less than the length of the array, increasing i by 1 each time.

2. Step 2: Analyze the print statement inside the loop

   Each iteration prints the element at index i, so all elements from start to end are printed in order.

3. Final Answer:

   Prints each element of the array from start to end -> Option A

4. Quick Check:

   Loop from 0 to len(arr)-1 prints all elements [OK]

Hint: Loop from 0 to len(arr) prints all elements in order [OK]

Common Mistakes:

• Thinking it prints in reverse order
• Assuming it prints only one element
• Confusing loop condition causing error

2. Which of the following is the correct syntax to traverse an array arr using a range-based for loop in Go?

easy

A. for value in arr { fmt.Println(value) }

B. for i := 0; i < len(arr); i++ { fmt.Println(arr[i]) }

C. for index, value := range arr { fmt.Println(index, value) }

D. foreach i in arr { fmt.Println(arr[i]) }

Solution

1. Step 1: Recall Go's range syntax

   Go uses for index, value := range arr to iterate over arrays or slices.

2. Step 2: Check each option

   for index, value := range arr { fmt.Println(index, value) } matches the correct syntax. for i := 0; i < len(arr); i++ { fmt.Println(arr[i]) } is valid but not range-based. Options C and D are invalid Go syntax.

3. Final Answer:

   for index, value := range arr { fmt.Println(index, value) } -> Option C

4. Quick Check:

   Range loop syntax = for index, value := range arr { fmt.Println(index, value) } [OK]

Hint: Use 'for index, value := range arr' for range loops in Go [OK]

Common Mistakes:

• Using Python or other language syntax
• Confusing traditional for loop with range loop
• Using invalid keywords like foreach

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

arr := []int{10, 20, 30, 40}
for i := len(arr) - 1; i >= 0; i-- {
    fmt.Print(arr[i], " ")
}

medium

A. 10 20 30 40

B. 40 30 20 10

C. Runtime error due to invalid index

D. No output

Solution

1. Step 1: Understand the loop range

   The loop starts at index len(arr)-1 (3) and decrements i until it reaches 0.

2. Step 2: Trace the printed values

   Indexes printed: 3, 2, 1, 0 corresponding to values 40, 30, 20, 10 printed with spaces.

3. Final Answer:

   40 30 20 10 -> Option B

4. Quick Check:

   Reverse loop prints elements backward [OK]

Hint: Loop from len(arr)-1 down to 0 prints array in reverse [OK]

Common Mistakes:

• Assuming forward order output
• Off-by-one errors causing runtime error
• Confusing loop condition direction

4. Identify the error in this Go code snippet that tries to print all elements of an array arr:

for i := 0; i <= len(arr); i++ {
    fmt.Println(arr[i])
}

medium

A. The print statement syntax is incorrect

B. The loop variable should start from 1 instead of 0

C. The array arr is not declared

D. The loop condition should be i < len(arr) to avoid index out of range

Solution

1. Step 1: Analyze the loop condition

   The loop runs while i <= len(arr), which means i can equal len(arr), an invalid index.

2. Step 2: Understand array indexing

   Array indexes go from 0 to len(arr)-1. Accessing arr[len(arr)] causes runtime error.

3. Final Answer:

   The loop condition should be i < len(arr) to avoid index out of range -> Option D

4. Quick Check:

   Index must be less than length, not less or equal [OK]

Hint: Use i < len(arr) to avoid out-of-range errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Starting loop at 1 instead of 0
• Ignoring array declaration errors

5. Given an array arr := []int{1, 2, 3, 4, 5}, which code snippet correctly prints only the elements at even indexes (0, 2, 4)?

hard

A. for i := 0; i < len(arr); i += 2 { fmt.Print(arr[i], " ") }

B. for i := 1; i < len(arr); i += 2 { fmt.Print(arr[i], " ") }

C. for _, v := range arr { if v%2 == 0 { fmt.Print(v, " ") } }

D. for i := len(arr)-1; i >= 0; i-- { if i%2 == 1 { fmt.Print(arr[i], " ") } }

Solution

1. Step 1: Understand the requirement

   We want to print elements at even indexes: 0, 2, 4.

2. Step 2: Check each option

   for i := 0; i < len(arr); i += 2 { fmt.Print(arr[i], " ") } increments i by 2 starting at 0, correctly accessing even indexes. for i := 1; i < len(arr); i += 2 { fmt.Print(arr[i], " ") } accesses odd indexes. for _, v := range arr { if v%2 == 0 { fmt.Print(v, " ") } } prints even values, not indexes. for i := len(arr)-1; i >= 0; i-- { if i%2 == 1 { fmt.Print(arr[i], " ") } } prints odd indexes in reverse.

3. Final Answer:

   for i := 0; i < len(arr); i += 2 { fmt.Print(arr[i], " ") } -> Option A

4. Quick Check:

   Increment by 2 from 0 prints even indexes [OK]

Hint: Increment index by 2 starting at 0 for even indexes [OK]

Common Mistakes:

• Confusing even indexes with even values
• Starting loop at 1 instead of 0
• Using incorrect loop increments