Radix Sort Algorithm in DSA Go - Time & Space Complexity

Time Complexity: Radix Sort Algorithm

O(d * n)

Understanding Time Complexity

We want to understand how the time taken by Radix Sort changes as the input size grows.

Specifically, how the number of digits and elements affect the work done.

Scenario Under Consideration

Analyze the time complexity of the following Radix Sort code snippet.

func radixSort(arr []int) []int {
    max := arr[0]
    for _, v := range arr {
        if v > max {
            max = v
        }
    }

    exp := 1
    n := len(arr)
    output := make([]int, n)

    for max/exp > 0 {
        count := make([]int, 10)
        for i := 0; i < n; i++ {
            count[(arr[i]/exp)%10]++
        }
        for i := 1; i < 10; i++ {
            count[i] += count[i-1]
        }
        for i := n - 1; i >= 0; i-- {
            output[count[(arr[i]/exp)%10]-1] = arr[i]
            count[(arr[i]/exp)%10]--
        }
        copy(arr, output)
        exp *= 10
    }
    return arr
}
    

This code sorts integers by processing digits from least to most significant.

Identify Repeating Operations

Look for loops and repeated steps:

• Primary operation: Counting sort on each digit place.
• How many times: Once for each digit of the largest number.
• Inside each digit pass, loops run over all elements multiple times.

How Execution Grows With Input

The work depends on two things: number of elements (n) and number of digits (d).

Input Size (n)	Digits (d)	Approx. Operations
10	2	About 20 loops over elements
100	3	About 300 loops over elements
1000	4	About 4000 loops over elements

Pattern observation: Operations grow roughly by n times d, so doubling n or d roughly doubles work.

Final Time Complexity

Time Complexity: O(d * n)

This means the time grows linearly with the number of elements and the number of digits in the largest number.

Common Mistake

[X] Wrong: "Radix Sort always runs in linear time O(n)."

[OK] Correct: The number of digits (d) matters; if numbers are very large, d grows and so does time.

Interview Connect

Understanding Radix Sort's time complexity helps you explain when it is efficient and when it might not be.

This skill shows you can analyze algorithms beyond simple loops, a key part of problem solving.

Self-Check

"What if we changed the base from 10 to 2 (binary digits)? How would the time complexity change?"

Practice

1. What is the main idea behind the Radix Sort Algorithm?

easy

A. Sort numbers by dividing the list into halves recursively

B. Sort numbers by processing individual digits from least to most significant

C. Sort numbers by comparing each pair directly

D. Sort numbers by swapping adjacent elements repeatedly

Solution

1. Step 1: Understand Radix Sort approach

   Radix Sort sorts numbers by grouping digits and sorting them digit by digit, starting from the least significant digit.

2. Step 2: Compare with other sorting methods

   Unlike comparison-based sorts (like bubble or merge sort), Radix Sort processes digits individually without direct element comparisons.

3. Final Answer:

   Sort numbers by processing individual digits from least to most significant -> Option B

4. Quick Check:

   Radix Sort = Digit-wise sorting [OK]

Hint: Radix Sort sorts digits from right to left [OK]

Common Mistakes:

• Thinking Radix Sort compares elements directly
• Confusing Radix Sort with divide and conquer methods
• Assuming it swaps adjacent elements like bubble sort

2. Which of the following is the correct way to initialize a slice of buckets for Radix Sort in Go?

easy

A. buckets := [][]int{}

B. buckets := make([]int, 10)

C. buckets := make([][]int, 10)

D. buckets := [10]int{}

Solution

1. Step 1: Identify bucket type needed

   Radix Sort uses buckets to hold multiple integers per digit, so each bucket must be a slice of ints.

2. Step 2: Choose correct slice initialization

   Using make([][]int, 10) creates a slice of 10 slices, perfect for buckets 0-9.

3. Final Answer:

   buckets := make([][]int, 10) -> Option C

4. Quick Check:

   Slice of slices for buckets = buckets := make([][]int, 10) [OK]

Hint: Buckets hold slices, so use [][]int with make [OK]

Common Mistakes:

• Using single int slice instead of slice of slices
• Using array instead of slice for dynamic buckets
• Initializing empty slice without length

3. What will be the output of the following Go code snippet using Radix Sort on the array [170, 45, 75, 90, 802, 24, 2, 66]?

package main
import "fmt"
func main() {
  arr := []int{170, 45, 75, 90, 802, 24, 2, 66}
  radixSort(arr)
  fmt.Println(arr)
}

func radixSort(arr []int) {
  max := getMax(arr)
  exp := 1
  for max/exp > 0 {
    countSort(arr, exp)
    exp *= 10
  }
}

func getMax(arr []int) int {
  max := arr[0]
  for _, v := range arr {
    if v > max {
      max = v
    }
  }
  return max
}

func countSort(arr []int, exp int) {
  n := len(arr)
  output := make([]int, n)
  count := make([]int, 10)

  for i := 0; i < n; i++ {
    index := (arr[i] / exp) % 10
    count[index]++
  }

  for i := 1; i < 10; i++ {
    count[i] += count[i-1]
  }

  for i := n - 1; i >= 0; i-- {
    index := (arr[i] / exp) % 10
    output[count[index]-1] = arr[i]
    count[index]--
  }

  for i := 0; i < n; i++ {
    arr[i] = output[i]
  }
}

medium

A. [2 24 45 66 75 90 802 170]

B. [802 170 90 75 66 45 24 2]

C. [170 45 75 90 802 24 2 66]

D. [2 24 45 66 75 90 170 802]

Solution

1. Step 1: Understand Radix Sort process on given array

   Radix Sort sorts by digits from least significant to most significant. After sorting by units, tens, and hundreds place, the array becomes sorted.

2. Step 2: Trace final sorted array

   The final sorted array is ascending: [2, 24, 45, 66, 75, 90, 170, 802].

3. Final Answer:

   [2 24 45 66 75 90 170 802] -> Option D

4. Quick Check:

   Radix Sort output = Sorted ascending array [OK]

Hint: Radix Sort outputs sorted array ascending [OK]

Common Mistakes:

• Confusing order of digits processed
• Misreading output as original array
• Mixing up output with descending order

4. Identify the bug in this Radix Sort countSort function snippet in Go:

func countSort(arr []int, exp int) {
  n := len(arr)
  output := make([]int, n)
  count := make([]int, 10)

  for i := 0; i < n; i++ {
    index := (arr[i] / exp) % 10
    count[index]++
  }

  for i := 1; i < 10; i++ {
    count[i] += count[i-1]
  }

  for i := 0; i < n; i++ {
    index := (arr[i] / exp) % 10
    output[count[index]-1] = arr[i]
    count[index]--
  }

  for i := 0; i < n; i++ {
    arr[i] = output[i]
  }
}

medium

A. The last for loop should iterate backwards from n-1 to 0

B. The count array size should be 100 instead of 10

C. The output slice should be initialized with zeros explicitly

D. The index calculation should use modulo 100 instead of 10

Solution

1. Step 1: Check order of placing elements in output

   To maintain stable sorting, elements must be placed from right to left (backwards) in output.

2. Step 2: Identify incorrect loop direction

   The code uses a forward loop (0 to n-1) which breaks stability; it should loop from n-1 down to 0.

3. Final Answer:

   The last for loop should iterate backwards from n-1 to 0 -> Option A

4. Quick Check:

   Stable sort needs backward loop [OK]

Hint: Place elements backward to keep stable order [OK]

Common Mistakes:

• Looping forward breaks stability
• Changing count array size unnecessarily
• Misunderstanding index calculation

5. You want to sort a list of strings representing numbers with varying lengths using Radix Sort in Go. Which approach correctly adapts Radix Sort for this case?

hard

A. Pad all strings with leading zeros to equal length, then sort digit by digit from right to left

B. Sort strings by length first, then apply Radix Sort on digits from left to right

C. Convert strings to integers and apply normal Radix Sort directly

D. Use Radix Sort only on the first character of each string

Solution

1. Step 1: Handle varying string lengths

   Pad strings with leading zeros so all have equal length, ensuring digit positions align.

2. Step 2: Apply Radix Sort digit by digit from right to left

   After padding, sort from least significant digit (rightmost) to most significant (leftmost) to maintain correct order.

3. Final Answer:

   Pad all strings with leading zeros to equal length, then sort digit by digit from right to left -> Option A

4. Quick Check:

   Padding equalizes length for Radix Sort [OK]

Hint: Pad strings to same length before sorting digits [OK]

Common Mistakes:

• Sorting by length first breaks digit alignment
• Converting strings to int may overflow or lose leading zeros
• Sorting only first character ignores full number