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DSA Goprogramming~15 mins

Array Insertion at Beginning in DSA Go - Build from Scratch

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Array Insertion at Beginning
📖 Scenario: You are managing a list of daily tasks. Sometimes, you get an urgent task that needs to be done first. You want to add this urgent task at the beginning of your task list.
🎯 Goal: Build a program that inserts a new task at the beginning of an existing array of tasks.
📋 What You'll Learn
Create an array of strings called tasks with these exact values: "email client", "write report", "team meeting"
Create a string variable called urgentTask with the value "fix bug"
Insert the urgentTask at the beginning of the tasks array
Print the updated tasks array
💡 Why This Matters
🌍 Real World
Managing task lists or queues where urgent items need to be prioritized at the front.
💼 Career
Understanding how to manipulate arrays and slices is essential for software development, especially in backend and systems programming.
Progress0 / 4 steps
1
Create the initial array of tasks
Create an array of strings called tasks with these exact values: "email client", "write report", "team meeting"
DSA Go
Hint

Use tasks := []string{} to create the array with the exact values inside curly braces.

2
Create the urgent task variable
Create a string variable called urgentTask with the value "fix bug"
DSA Go
Hint

Use urgentTask := "fix bug" to create the string variable.

3
Insert the urgent task at the beginning
Insert the urgentTask at the beginning of the tasks array by creating a new array with urgentTask followed by the existing tasks
DSA Go
Hint

Use append with a new slice containing urgentTask and then the existing tasks using ... to expand.

4
Print the updated tasks array
Print the updated tasks array using fmt.Println(tasks)
DSA Go
Hint

Use fmt.Println(tasks) to print the array.

Practice

(1/5)
1. What happens when you insert an element at the beginning of an array in Go?
easy
A. The array size decreases by one.
B. The new element replaces the first element without shifting others.
C. All existing elements shift one position to the right to make space.
D. The new element is added at the end of the array.

Solution

  1. Step 1: Understand array behavior in Go

    Arrays have fixed size, so to insert at the beginning, elements must shift right.
  2. Step 2: Effect of insertion at beginning

    The first element moves to second position, second to third, and so on, making space at index 0.
  3. Final Answer:

    All existing elements shift one position to the right to make space. -> Option C
  4. Quick Check:

    Insertion at start shifts elements right = A [OK]
Hint: Insertion at start shifts all elements right by one [OK]
Common Mistakes:
  • Thinking new element replaces first without shifting
  • Assuming array size changes automatically
  • Adding element at the end instead of beginning
2. Which of the following Go code snippets correctly inserts an element `x` at the beginning of a slice `s`?
easy
A. s = s[1:]
B. s = append(s, x)
C. s[0] = x
D. s = append([]int{x}, s...)

Solution

  1. Step 1: Understand slice append syntax

    Appending a slice to another uses append(slice1, slice2...), so to add at start, prepend with new slice.
  2. Step 2: Analyze each option

    s = append([]int{x}, s...) prepends x correctly; B appends at end; C replaces first element; D slices off first element.
  3. Final Answer:

    s = append([]int{x}, s...) -> Option D
  4. Quick Check:

    Prepend with append([]int{x}, s...) = A [OK]
Hint: Use append with new slice and ... to insert at start [OK]
Common Mistakes:
  • Using append(s, x) adds at end, not start
  • Replacing element instead of inserting
  • Slicing removes elements instead of adding
3. What is the output of this Go code?
package main
import "fmt"
func main() {
  s := []int{2, 3, 4}
  s = append([]int{1}, s...)
  fmt.Println(s)
}
medium
A. [1 2 3 4]
B. [2 3 4 1]
C. [1 3 4]
D. [2 3 4]

Solution

  1. Step 1: Initial slice and append operation

    Initial slice s is [2 3 4]. append([]int{1}, s...) adds 1 at start.
  2. Step 2: Resulting slice after append

    New slice becomes [1 2 3 4] because 1 is prepended before all elements.
  3. Final Answer:

    [1 2 3 4] -> Option A
  4. Quick Check:

    Prepending 1 to [2 3 4] = [1 2 3 4] [OK]
Hint: append([]int{x}, s...) prepends x to slice s [OK]
Common Mistakes:
  • Thinking append adds at end by default
  • Ignoring ... operator to expand slice
  • Confusing slice contents after append
4. Identify the error in this Go code that tries to insert 0 at the beginning of slice `s`:
package main
import "fmt"
func main() {
  s := []int{1, 2, 3}
  s[0] = 0
  fmt.Println(s)
}
medium
A. It replaces the first element instead of inserting a new one.
B. It causes a runtime panic due to index out of range.
C. It appends 0 at the end instead of the beginning.
D. It creates a new slice with 0 at the start.

Solution

  1. Step 1: Analyze the assignment s[0] = 0

    This replaces the value at index 0 with 0, not inserting a new element.
  2. Step 2: Effect on slice length and contents

    Slice length stays same; no shifting occurs; original first element lost.
  3. Final Answer:

    It replaces the first element instead of inserting a new one. -> Option A
  4. Quick Check:

    Assignment replaces element, does not insert [OK]
Hint: Assignment replaces element; use append to insert [OK]
Common Mistakes:
  • Thinking s[0] = 0 inserts element
  • Expecting slice length to increase
  • Confusing replace with insert
5. Given a slice `s := []int{3, 4, 5}`, write Go code to insert the numbers 1 and 2 at the beginning so that the final slice is `[1 2 3 4 5]`. Which code snippet achieves this correctly?
hard
A. s[0], s[1] = 1, 2
B. s = append([]int{1, 2}, s...)
C. s = append(s[:0], 1, 2)
D. s = append(s, 1, 2)

Solution

  1. Step 1: Understand slice prepending with append

    To add multiple elements at start, create a new slice with them and append original slice after.
  2. Step 2: Analyze options

    s = append([]int{1, 2}, s...) prepends [1, 2] correctly; A replaces first two elements; B appends at end; C overwrites start without preserving rest.
  3. Final Answer:

    s = append([]int{1, 2}, s...) -> Option B
  4. Quick Check:

    Prepend multiple elements with append([]int{...}, s...) [OK]
Hint: Use append with slice of new elements and ... to prepend [OK]
Common Mistakes:
  • Appending instead of prepending
  • Overwriting elements without shifting
  • Misusing slice syntax causing data loss
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