radix sort code in c++ code example

Example 1: Radix Sort in c++

#include <iostream> 
using namespace std; 
  

int getMax(int arr[], int n) 
{ 
    int mx = arr[0]; 
    for (int i = 1; i < n; i++) 
        if (arr[i] > mx) 
            mx = arr[i]; 
    return mx; 
} 
  

void countSort(int arr[], int n, int exp) 
{ 
    int output[n]; 
    int i, count[10] = { 0 }; 
  
    
    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]--; 
    } 
  
    for (i = 0; i < n; i++) 
        arr[i] = output[i]; 
} 
  


void radixsort(int arr[], int n) 
{ 

    int m = getMax(arr, n); 
  
   
    for (int exp = 1; m / exp > 0; exp *= 10) 
        countSort(arr, n, exp); 
} 
  

void print(int arr[], int n) 
{ 
    for (int i = 0; i < n; i++) 
        cout << arr[i] << " "; 
} 
  

int main() 
{ 
    int arr[] = { 170, 45, 75, 90, 802, 24, 2, 66 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
      

      radixsort(arr, n); 
    print(arr, n); 
    return 0; 
}

Example 2: radix sort

function getDigit(num, i) {
  return Math.floor(Math.abs(num) / Math.pow(10,i)) % 10
}

function digitCount(num) {
  if(num === 0 ) return 1
  return Math.floor(Math.log10(Math.abs(num))) + 1
}

function mostDigitCount(nums) {
  let maxDigit = 0
  for(let i = 0;i< nums.length;i++) {
    maxDigit = Math.max(maxDigit, digitCount(nums[i]))
  }
  return maxDigit
}

function Radix(nums){
  let maxDigitCount = mostDigitCount(nums)
  for(let k=0; k< maxDigitCount; k++) {
    let digitbucket = Array.from({length:10} , () => [])
    for(let i=0; i< nums.length; i++) {
      let digit = getDigit(nums[i], k)
      digitbucket[digit].push(nums[i])
    }
    nums = [].concat(...digitbucket)
  }
  return nums
}

console.log(Radix([23,123, 23333,444444,55555555]))

Example 3: radix sort pseudocode

Radix-Sort(A, d)
       for j = 1 to d do
            int count[10] = {0};
            for i = 0 to n do
                count[key of(A[i]) in pass j]++
            for k = 1 to 10 do
                count[k] = count[k] + count[k-1]
            for i = n-1 downto 0 do
                result[ count[key of(A[i])] ] = A[j]
                count[key of(A[i])]--
            for i=0 to n do
                A[i] = result[i]
       end for(j)
 end func