How to efficiently sort a big list dates in 20's

How to efficiently sort a big list dates in 20’s

Difficulty Level : Medium
Last Updated : 29 Sep, 2020

Given a big list of dates in ’20s, how to efficiently sort the list.
Example:

Input:
       Date arr[] = {{20,  1, 2014},
                    {25,  3, 2010},
                    { 3, 12, 2000},
                    {18, 11, 2001},
                    {19,  4, 2015},
                    { 9,  7, 2005}}
Output:
      Date arr[] = {{ 3, 12, 2000},
                    {18, 11, 2001},
                    { 9,  7, 2005},
                    {25,  3, 2010},
                    {20,  1, 2014},
                    {19,  4, 2015}}

A Simple Solution is to use an O(N log N) algorithm like Merge Sort and a custom comparator. But we can sort the list in O(n) time using Radix Sort. In a typical Radix Sort implementation, we first sort by the last digit, then by second last digit, and so on. It is highly recommended to first go through the radix sort to understand this method. In this method we sort in the following order:

First sort by day using counting sort
Then sort by month using counting sort
Finally, sort by year using counting sort

As the number of days, months, and years are fixed, all three steps take O(n) time. Therefore, the overall time complexity is O(n).

Below is the C++ implementation of the above idea:

C++

// C++ program to sort an array
// of dates using Radix Sort
 
#include <bits/stdc++.h>
using namespace std;
 
// Utilitiy function to obtain 
// maximum date or month or year
int getMax(int arr[][3],int n, int q)
{
    int maxi = INT_MIN;
    for(int i=0;i<n;i++){
        maxi = max(maxi,arr[i][q]);
    }
    return maxi;
}
 
// A function to do counting sort of arr[]
// according to given q i.e.
// (0 for day, 1 for month, 2 for year)
void sortDatesUtil(int arr[][3], 
                   int n, int q)
{
    int maxi = getMax(arr,n,q);
    int p = 1;
    while(maxi>0){
        //to extract last digit  divide 
        // by p then %10
        // Store count of occurrences in cnt[]
        int cnt[10]={0};
         
        for(int i=0;i<n;i++)
        {
            cnt[(arr[i][q]/p)%10]++;
        }
        for(int i=1;i<10;i++)
        {
            cnt[i]+=cnt[i-1];
        }
        int ans[n][3];
        for(int i=n-1;i>=0;i--)
        {
            int lastDigit = (arr[i][q]/p)%10;
             
            // Build the output array
            for(int j=0;j<3;j++)
            {
                ans[cnt[lastDigit]-1][j]
                  =arr[i][j];    
            }
            cnt[lastDigit]--;
        }
        // Copy the output array to arr[],
        // so that arr[] now
        // contains sorted numbers
        // according to current digit
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<3;j++)
            {
                arr[i][j] = ans[i][j];
            }
        }
        // update p to get 
        // the next digit
        p*=10;
        maxi/=10;
    }
}
 
// The main function that sorts
// array of dates
// using Radix Sort
void sortDates(int dates[][3], int n)
{
    // First sort by day
    sortDatesUtil(dates, n, 0);
   
    // Then by month
    sortDatesUtil(dates, n, 1); 
    // Finally by year
    sortDatesUtil(dates, n, 2);
}
 
// A utility function to print an array
void printArr(int arr[][3], int n)
{
   for(int i=0;i<6;i++)
   {
        for(int j=0;j<3;j++)
        {
            cout<<arr[i][j]<<" ";
        }
        cout<<endl;
    }
}
 
// Driver Code
int main() 
{ 
    int dates[][3] = {{20,  1, 2014},
                    {25,  3, 2010},
                    { 3, 12, 2000},
                    {18, 11, 2000},
                    {19,  4, 2015},
                    { 9,  7, 2005}};
    int n = sizeof(dates)/sizeof(dates[0]);
   
    // Function Call
    sortDates(dates,n);
    cout<<"\nSorted Dates\n";
    printArr(dates,n);
    return 0; 
} 

Output

Sorted Dates
18 11 2000 
3 12 2000 
9 7 2005 
25 3 2010 
20 1 2014 
19 4 2015

This article is contributed by tpriyanshu. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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