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

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  • Difficulty Level : Expert
  • Last Updated : 27 Aug, 2022
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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

 

Time Complexity: O(n)

Auxiliary Space: O(n), since n extra space has been taken.


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