Discover how sorting by digits can turn a huge mess into a neat, quick order!
Why Radix Sort Algorithm in DSA Go?
Imagine you have a huge stack of mail envelopes with different zip codes, and you need to sort them by their zip codes manually.
You try to compare each zip code digit by digit, but it takes forever and you keep mixing them up.
Sorting by comparing whole numbers or strings one by one is slow and confusing when the list is very long.
It's easy to make mistakes and waste time rechecking and swapping items repeatedly.
Radix Sort sorts numbers digit by digit, starting from the smallest digit to the largest.
This way, it groups numbers by each digit quickly without comparing whole numbers directly.
It's like sorting mail by the last digit of the zip code first, then the next digit, and so on, making the process smooth and fast.
for i := 0; i < len(numbers); i++ { for j := i + 1; j < len(numbers); j++ { if numbers[i] > numbers[j] { numbers[i], numbers[j] = numbers[j], numbers[i] } } }
func radixSort(numbers []int) {
// Sort numbers digit by digit
// from least significant to most significant
}Radix Sort enables fast and reliable sorting of large lists of numbers without complex comparisons.
Sorting millions of phone numbers or social security numbers quickly in a database for fast searching and reporting.
Manual sorting by comparing whole numbers is slow and error-prone.
Radix Sort sorts numbers digit by digit, making sorting faster and simpler.
This method is great for sorting large lists of numbers efficiently.
Practice
Radix Sort Algorithm?Solution
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.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.Final Answer:
Sort numbers by processing individual digits from least to most significant -> Option BQuick Check:
Radix Sort = Digit-wise sorting [OK]
- Thinking Radix Sort compares elements directly
- Confusing Radix Sort with divide and conquer methods
- Assuming it swaps adjacent elements like bubble sort
Solution
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.Step 2: Choose correct slice initialization
Usingmake([][]int, 10)creates a slice of 10 slices, perfect for buckets 0-9.Final Answer:
buckets := make([][]int, 10) -> Option CQuick Check:
Slice of slices for buckets = buckets := make([][]int, 10) [OK]
- Using single int slice instead of slice of slices
- Using array instead of slice for dynamic buckets
- Initializing empty slice without length
[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]
}
}Solution
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.Step 2: Trace final sorted array
The final sorted array is ascending: [2, 24, 45, 66, 75, 90, 170, 802].Final Answer:
[2 24 45 66 75 90 170 802] -> Option DQuick Check:
Radix Sort output = Sorted ascending array [OK]
- Confusing order of digits processed
- Misreading output as original array
- Mixing up output with descending order
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]
}
}Solution
Step 1: Check order of placing elements in output
To maintain stable sorting, elements must be placed from right to left (backwards) in output.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.Final Answer:
The last for loop should iterate backwards from n-1 to 0 -> Option AQuick Check:
Stable sort needs backward loop [OK]
- Looping forward breaks stability
- Changing count array size unnecessarily
- Misunderstanding index calculation
Solution
Step 1: Handle varying string lengths
Pad strings with leading zeros so all have equal length, ensuring digit positions align.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.Final Answer:
Pad all strings with leading zeros to equal length, then sort digit by digit from right to left -> Option AQuick Check:
Padding equalizes length for Radix Sort [OK]
- Sorting by length first breaks digit alignment
- Converting strings to int may overflow or lose leading zeros
- Sorting only first character ignores full number
Ready to prove it?
Pass the test and this pattern moves unknown
