Skip to content
Related Articles

Related Articles

Improve Article

Count of sub-sets of size n with total element sum divisible by 3

  • Difficulty Level : Expert
  • Last Updated : 01 Jun, 2021

Given an integer n and a range [l, r], the task is to find the count of total sub-sets of size n with integers from the given range such that the total sum of its elements is divisible by 3.
Examples: 
 

Input: n = 2, l = 1, r = 5 
Output:
Possible sub-sets are {1, 2}, {2, 1}, {3, 3}, {5, 1}, {1, 5}, {4, 2}, {2, 4}, {5, 4} and {4, 5} 
Input: n = 3, l = 9, r = 9 
Output:
{9, 9, 9} is the only possible sub-set 
 

 

Approach: Since we need the sum of the sub-set elements to be divisible by 3. So, instead of caring about the numbers, we will count the numbers such that they give remainder 0, 1 and 2 on dividing with 3 separately by the formula given below: 
 

For example, an element k such that k % 3 = 2 can be found as k = 3 * x + 2 for some integer x
Then we have l ≤ (3 * x) + 2 ≤ r 
l – 2 ≤ (3 * x) ≤ r – 2 
ceil((l – 2) / 3) ≤ x ≤ floor((r – 2) / 3) 
 



Now, by dynamic programming dp[i][j] we can check how many elements will give a sum that is divisible by 3. Here dp[i][j] represents the sum of first i elements that give remainder j on dividing by 3.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
#define MOD 1000000007
#define ll long long int
using namespace std;
 
// Function to return the total number of
// required sub-sets
int totalSubSets(ll n, ll l, ll r)
{
 
    // Variable to store total elements
    // which on dividing by 3  give
    // remainder 0, 1 and 2 respectively
    ll zero = floor((double)r / 3)
              - ceil((double)l / 3) + 1;
    ll one = floor((double)(r - 1) / 3)
             - ceil((double)(l - 1) / 3) + 1;
    ll two = floor((double)(r - 2) / 3)
             - ceil((double)(l - 2) / 3) + 1;
 
    // Create a dp table
    ll dp[n][3];
    memset(dp, 0, sizeof(dp));
    dp[0][0] = zero;
    dp[0][1] = one;
    dp[0][2] = two;
 
    // Process for n states and store
    // the sum (mod 3) for 0, 1 and 2
    for (ll i = 1; i < n; ++i) {
 
        // Use of MOD for large numbers
        dp[i][0] = ((dp[i - 1][0] * zero)
                    + (dp[i - 1][1] * two)
                    + (dp[i - 1][2] * one))
                   % MOD;
        dp[i][1] = ((dp[i - 1][0] * one)
                    + (dp[i - 1][1] * zero)
                    + (dp[i - 1][2] * two))
                   % MOD;
        dp[i][2] = ((dp[i - 1][0] * two)
                    + (dp[i - 1][1] * one)
                    + (dp[i - 1][2] * zero))
                   % MOD;
    }
 
    // Final answer store at dp[n - 1][0]
    return dp[n - 1][0];
}
 
// Driver Program
int main()
{
    ll n = 5;
    ll l = 10;
    ll r = 100;
    cout << totalSubSets(n, l, r);
    return 0;
}

Java




// Java implementation of the approach
 
class GFG
{
         
    static int MOD = 1000000007;
     
    // Function to return the total number of
    // required sub-sets
    static int totalSubSets(int n, int l, int r)
    {
     
        // Variable to store total elements
        // which on dividing by 3 give
        // remainder 0, 1 and 2 respectively
        int zero = (int)Math.floor((double)r / 3)
                - (int)Math.ceil((double)l / 3) + 1;
        int one = (int)Math.floor((double)(r - 1) / 3)
                - (int)Math.ceil((double)(l - 1) / 3) + 1;
        int two = (int)Math.floor((double)(r - 2) / 3)
                - (int)Math.ceil((double)(l - 2) / 3) + 1;
     
        // Create a dp table
        int [][] dp = new int[n][3];
     
        dp[0][0] = zero;
        dp[0][1] = one;
        dp[0][2] = two;
     
        // Process for n states and store
        // the sum (mod 3) for 0, 1 and 2
        for (int i = 1; i < n; ++i)
        {
     
            // Use of MOD for large numbers
            dp[i][0] = ((dp[i - 1][0] * zero)
                        + (dp[i - 1][1] * two)
                        + (dp[i - 1][2] * one))
                    % MOD;
            dp[i][1] = ((dp[i - 1][0] * one)
                        + (dp[i - 1][1] * zero)
                        + (dp[i - 1][2] * two))
                    % MOD;
            dp[i][2] = ((dp[i - 1][0] * two)
                        + (dp[i - 1][1] * one)
                        + (dp[i - 1][2] * zero))
                    % MOD;
        }
     
        // Final answer store at dp[n - 1][0]
        return dp[n - 1][0];
    }
     
    // Driver Program
    public static void main(String []args)
    {
        int n = 5;
        int l = 10;
        int r = 100;
        System.out.println(totalSubSets(n, l, r));
    }
}
 
// This code is contributed by ihritik

Python3




# Python3 implementation of the approach
import math
 
# Function to return the total
# number of required sub-sets
def totalSubSets(n, l, r):
     
    MOD = 1000000007 ;
     
    # Variable to store total elements
    # which on dividing by 3 give
    # remainder 0, 1 and 2 respectively
    zero = (math.floor(r / 3) -
            math.ceil(l / 3) + 1);
    one = (math.floor((r - 1) / 3) -
           math.ceil((l - 1) / 3) + 1);
    two = (math.floor((r - 2) / 3) -
           math.ceil((l - 2) / 3) + 1);
 
    # Create a dp table
    dp = [[0 for x in range(3)]
             for y in range(n)]
             
    dp[0][0] = zero;
    dp[0][1] = one;
    dp[0][2] = two;
 
    # Process for n states and store
    # the sum (mod 3) for 0, 1 and 2
    for i in range(1, n):
 
        # Use of MOD for large numbers
        dp[i][0] = ((dp[i - 1][0] * zero) +
                    (dp[i - 1][1] * two) +
                    (dp[i - 1][2] * one)) % MOD;
        dp[i][1] = ((dp[i - 1][0] * one) +
                    (dp[i - 1][1] * zero) +
                    (dp[i - 1][2] * two)) % MOD;
        dp[i][2] = ((dp[i - 1][0] * two)+
                    (dp[i - 1][1] * one) +
                    (dp[i - 1][2] * zero)) % MOD;
 
    # Final answer store at dp[n - 1][0]
    return dp[n - 1][0];
 
# Driver Code
n = 5;
l = 10;
r = 100;
print(totalSubSets(n, l, r));
     
# This code is contributed
# by chandan_jnu

C#




// C# implementation of the approach
using System;
 
class GFG
{
         
    static int MOD = 1000000007;
     
    // Function to return the total number of
    // required sub-sets
    static int totalSubSets(int n, int l, int r)
    {
     
        // Variable to store total elements
        // which on dividing by 3 give
        // remainder 0, 1 and 2 respectively
        int zero = (int)Math.Floor((double)r / 3)
                - (int)Math.Ceiling((double)l / 3) + 1;
        int one = (int)Math.Floor((double)(r - 1) / 3)
                - (int)Math.Ceiling((double)(l - 1) / 3) + 1;
        int two = (int)Math.Floor((double)(r - 2) / 3)
                - (int)Math.Ceiling((double)(l - 2) / 3) + 1;
     
        // Create a dp table
        int [, ] dp = new int[n, 3];
     
        dp[0,0] = zero;
        dp[0,1] = one;
        dp[0,2] = two;
     
        // Process for n states and store
        // the sum (mod 3) for 0, 1 and 2
        for (int i = 1; i < n; ++i)
        {
     
            // Use of MOD for large numbers
            dp[i,0] = ((dp[i - 1, 0] * zero)
                        + (dp[i - 1, 1] * two)
                        + (dp[i - 1, 2] * one))
                    % MOD;
            dp[i,1] = ((dp[i - 1, 0] * one)
                        + (dp[i - 1, 1] * zero)
                        + (dp[i - 1, 2] * two))
                    % MOD;
            dp[i,2] = ((dp[i - 1, 0] * two)
                        + (dp[i - 1, 1] * one)
                        + (dp[i - 1, 2] * zero))
                    % MOD;
        }
     
        // Final answer store at dp[n - 1,0]
        return dp[n - 1, 0];
    }
     
    // Driver Program
    public static void Main()
    {
        int n = 5;
        int l = 10;
        int r = 100;
        Console.WriteLine(totalSubSets(n, l, r));
    }
}
 
// This code is contributed by ihritik

PHP




<?php
# Php implementation of the approach
 
# Function to return the total number of
# required sub-sets
function totalSubSets($n, $l, $r)
{
     
    $MOD = 1000000007 ;
    // Variable to store total elements
    // which on dividing by 3 give
    // remainder 0, 1 and 2 respectively
    $zero = floor($r / 3)
            - ceil($l / 3) + 1;
    $one = floor(($r - 1) / 3)
            - ceil(($l - 1) / 3) + 1;
    $two = floor(($r - 2) / 3)
            - ceil(($l - 2) / 3) + 1;
 
    // Create a dp table
    $dp = array() ;
    for($i = 0; $i < $n; $i++)
        for($j = 0; $j < 3; $j++)
            $dp[$i][$j] = 0 ;
             
    $dp[0][0] = $zero;
    $dp[0][1] = $one;
    $dp[0][2] = $two;
 
    // Process for n states and store
    // the sum (mod 3) for 0, 1 and 2
    for ($i = 1; $i < $n; ++$i)
    {
 
        // Use of MOD for large numbers
        $dp[$i][0] = (($dp[$i - 1][0] * $zero)
                    + ($dp[$i - 1][1] * $two)
                    + ($dp[$i - 1][2] * $one))
                % $MOD;
        $dp[$i][1] = (($dp[$i - 1][0] * $one)
                    + ($dp[$i - 1][1] * $zero)
                    + ($dp[$i - 1][2] * $two))
                % $MOD;
        $dp[$i][2] = (($dp[$i - 1][0] * $two)
                    + ($dp[$i - 1][1] * $one)
                    + ($dp[$i - 1][2] * $zero))
                % $MOD;
    }
 
    // Final answer store at dp[n - 1][0]
    return $dp[$n - 1][0];
}
 
// Driver Program
$n = 5;
$l = 10;
$r = 100;
echo totalSubSets($n, $l, $r);
     
// This code is contributed by Ryuga
?>

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    let MOD = 1000000007;
       
    // Function to return the total number of
    // required sub-sets
    function totalSubSets(n, l, r)
    {
       
        // Variable to store total elements
        // which on dividing by 3 give
        // remainder 0, 1 and 2 respectively
        let zero = Math.floor(r / 3)
                - Math.ceil(l / 3) + 1;
        let one = Math.floor((r - 1) / 3)
                - Math.ceil((l - 1) / 3) + 1;
        let two = Math.floor((r - 2) / 3)
                - Math.ceil((l - 2) / 3) + 1;
       
        // Create a dp table
        let dp = new Array(n);
        for(let i = 0; i < n; i++)
        {
            dp[i] = new Array(3);
            for(let j = 0; j < 3; j++)
            {
                dp[i][j] = 0;
            }
        }
       
        dp[0][0] = zero;
        dp[0][1] = one;
        dp[0][2] = two;
       
        // Process for n states and store
        // the sum (mod 3) for 0, 1 and 2
        for (let i = 1; i < n; ++i)
        {
       
            // Use of MOD for large numbers
            dp[i][0] = ((dp[i - 1][0] * zero)
                        + (dp[i - 1][1] * two)
                        + (dp[i - 1][2] * one))
                    % MOD;
            dp[i][1] = ((dp[i - 1][0] * one)
                        + (dp[i - 1][1] * zero)
                        + (dp[i - 1][2] * two))
                    % MOD;
            dp[i][2] = ((dp[i - 1][0] * two)
                        + (dp[i - 1][1] * one)
                        + (dp[i - 1][2] * zero))
                    % MOD;
        }
       
        // Final answer store at dp[n - 1][0]
        return dp[n - 1][0];
    }
     
    let n = 5;
    let l = 10;
    let r = 100;
    document.write(totalSubSets(n, l, r));
     
</script>
Output: 
80107136

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :