Given an unsorted integer (positive values only) array of size ‘n’, we can form a group of two or three, the group should be such that the sum of all elements in that group is a multiple of 3. Count all possible number of groups that can be generated in this way.
Examples:
Input: arr[] = {3, 6, 7, 2, 9} Output: 8 // Groups are {3,6}, {3,9}, {9,6}, {7,2}, {3,6,9}, // {3,7,2}, {7,2,6}, {7,2,9} Input: arr[] = {2, 1, 3, 4} Output: 4 // Groups are {2,1}, {2,4}, {2,1,3}, {2,4,3}
The idea is to see remainder of every element when divided by 3. A set of elements can form a group only if sum of their remainders is multiple of 3.
For example : 8, 4, 12. Now, the remainders are 2, 1, and 0 respectively. This means 8 is 2 distance away from 3s multiple (6), 4 is 1 distance away from 3s multiple(3), and 12 is 0 distance away. So, we can write the sum as 8 (can be written as 6+2), 4 (can be written as 3+1), and 12 (can be written as 12+0). Now the sum of 8, 4 and 12 can be written as 6+2+3+1+12+0. Now, 6+3+12 will always be divisible by 3 as all the terms are multiple of three. Now, we only need to check if 2+1+0 (remainders) is divisible by 3 or not which makes the complete sum divisible by 3.
Since the task is to enumerate groups, we count all elements with different remainders.
1. Hash all elements in a count array based on remainder, i.e, for all elements a[i], do c[a[i]%3]++; 2. Now c[0] contains the number of elements which when divided by 3 leave remainder 0 and similarly c[1] for remainder 1 and c[2] for 2. 3. Now for group of 2, we have 2 possibilities a. 2 elements of remainder 0 group. Such possibilities are c[0]*(c[0]-1)/2 b. 1 element of remainder 1 and 1 from remainder 2 group Such groups are c[1]*c[2]. 4. Now for group of 3,we have 4 possibilities a. 3 elements from remainder group 0. No. of such groups are c[0]C3 b. 3 elements from remainder group 1. No. of such groups are c[1]C3 c. 3 elements from remainder group 2. No. of such groups are c[2]C3 d. 1 element from each of 3 groups. No. of such groups are c[0]*c[1]*c[2]. 5. Add all the groups in steps 3 and 4 to obtain the result.
Implementation:
// C++ Program to count all possible // groups of size 2 or 3 that have // sum as multiple of 3 #include<bits/stdc++.h> using namespace std;
// Returns count of all possible groups // that can be formed from elements of a[]. int findgroups( int arr[], int n)
{ // Create an array C[3] to store counts
// of elements with remainder 0, 1 and 2.
// c[i] would store count of elements
// with remainder i
int c[3] = {0}, i;
int res = 0; // To store the result
// Count elements with remainder 0, 1 and 2
for (i=0; i<n; i++)
c[arr[i]%3]++;
// Case 3.a: Count groups of size 2
// from 0 remainder elements
res += ((c[0]*(c[0]-1))>>1);
// Case 3.b: Count groups of size 2 with
// one element with 1 remainder and other
// with 2 remainder
res += c[1] * c[2];
// Case 4.a: Count groups of size 3
// with all 0 remainder elements
res += (c[0] * (c[0]-1) * (c[0]-2))/6;
// Case 4.b: Count groups of size 3
// with all 1 remainder elements
res += (c[1] * (c[1]-1) * (c[1]-2))/6;
// Case 4.c: Count groups of size 3
// with all 2 remainder elements
res += ((c[2]*(c[2]-1)*(c[2]-2))/6);
// Case 4.c: Count groups of size 3
// with different remainders
res += c[0]*c[1]*c[2];
// Return total count stored in res
return res;
} // Driver Code int main()
{ int arr[] = {3, 6, 7, 2, 9};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << "Required number of groups are "
<< findgroups(arr,n) << endl;
return 0;
} // This code is contributed // by Akanksha Rai |
// C Program to count all possible // groups of size 2 or 3 that have // sum as multiple of 3 #include<stdio.h> // Returns count of all possible groups that can be formed from elements // of a[]. int findgroups( int arr[], int n)
{ // Create an array C[3] to store counts of elements with remainder
// 0, 1 and 2. c[i] would store count of elements with remainder i
int c[3] = {0}, i;
int res = 0; // To store the result
// Count elements with remainder 0, 1 and 2
for (i=0; i<n; i++)
c[arr[i]%3]++;
// Case 3.a: Count groups of size 2 from 0 remainder elements
res += ((c[0]*(c[0]-1))>>1);
// Case 3.b: Count groups of size 2 with one element with 1
// remainder and other with 2 remainder
res += c[1] * c[2];
// Case 4.a: Count groups of size 3 with all 0 remainder elements
res += (c[0] * (c[0]-1) * (c[0]-2))/6;
// Case 4.b: Count groups of size 3 with all 1 remainder elements
res += (c[1] * (c[1]-1) * (c[1]-2))/6;
// Case 4.c: Count groups of size 3 with all 2 remainder elements
res += ((c[2]*(c[2]-1)*(c[2]-2))/6);
// Case 4.c: Count groups of size 3 with different remainders
res += c[0]*c[1]*c[2];
// Return total count stored in res
return res;
} // Driver program to test above functions int main()
{ int arr[] = {3, 6, 7, 2, 9};
int n = sizeof (arr)/ sizeof (arr[0]);
printf ( "Required number of groups are %d\n" , findgroups(arr,n));
return 0;
} |
// Java Program to count all possible // groups of size 2 or 3 that have // sum as multiple of 3 class FindGroups
{ // Returns count of all possible groups that can be formed from elements
// of a[].
int findgroups( int arr[], int n)
{
// Create an array C[3] to store counts of elements with remainder
// 0, 1 and 2. c[i] would store count of elements with remainder i
int c[] = new int []{ 0 , 0 , 0 };
int i;
int res = 0 ; // To store the result
// Count elements with remainder 0, 1 and 2
for (i = 0 ; i < n; i++)
c[arr[i] % 3 ]++;
// Case 3.a: Count groups of size 2 from 0 remainder elements
res += ((c[ 0 ] * (c[ 0 ] - 1 )) >> 1 );
// Case 3.b: Count groups of size 2 with one element with 1
// remainder and other with 2 remainder
res += c[ 1 ] * c[ 2 ];
// Case 4.a: Count groups of size 3 with all 0 remainder elements
res += (c[ 0 ] * (c[ 0 ] - 1 ) * (c[ 0 ] - 2 )) / 6 ;
// Case 4.b: Count groups of size 3 with all 1 remainder elements
res += (c[ 1 ] * (c[ 1 ] - 1 ) * (c[ 1 ] - 2 )) / 6 ;
// Case 4.c: Count groups of size 3 with all 2 remainder elements
res += ((c[ 2 ] * (c[ 2 ] - 1 ) * (c[ 2 ] - 2 )) / 6 );
// Case 4.c: Count groups of size 3 with different remainders
res += c[ 0 ] * c[ 1 ] * c[ 2 ];
// Return total count stored in res
return res;
}
public static void main(String[] args)
{
FindGroups groups = new FindGroups();
int arr[] = { 3 , 6 , 7 , 2 , 9 };
int n = arr.length;
System.out.println( "Required number of groups are "
+ groups.findgroups(arr, n));
}
} |
# Python3 Program to Count groups # of size 2 or 3 that have sum # as multiple of 3 # Returns count of all possible # groups that can be formed # from elements of a[]. def findgroups(arr, n):
# Create an array C[3] to store
# counts of elements with
# remainder 0, 1 and 2. c[i]
# would store count of elements
# with remainder i
c = [ 0 , 0 , 0 ]
# To store the result
res = 0
# Count elements with remainder
# 0, 1 and 2
for i in range ( 0 , n):
c[arr[i] % 3 ] + = 1
# Case 3.a: Count groups of size
# 2 from 0 remainder elements
res + = ((c[ 0 ] * (c[ 0 ] - 1 )) >> 1 )
# Case 3.b: Count groups of size
# 2 with one element with 1
# remainder and other with 2 remainder
res + = c[ 1 ] * c[ 2 ]
# Case 4.a: Count groups of size
# 3 with all 0 remainder elements
res + = (c[ 0 ] * (c[ 0 ] - 1 ) * (c[ 0 ] - 2 )) / 6
# Case 4.b: Count groups of size 3
# with all 1 remainder elements
res + = (c[ 1 ] * (c[ 1 ] - 1 ) * (c[ 1 ] - 2 )) / 6
# Case 4.c: Count groups of size 3
# with all 2 remainder elements
res + = ((c[ 2 ] * (c[ 2 ] - 1 ) * (c[ 2 ] - 2 )) / 6 )
# Case 4.c: Count groups of size 3
# with different remainders
res + = c[ 0 ] * c[ 1 ] * c[ 2 ]
# Return total count stored in res
return res
# Driver program arr = [ 3 , 6 , 7 , 2 , 9 ]
n = len (arr)
print ( "Required number of groups are" ,
int (findgroups(arr, n)))
# This article is contributed by shreyanshi_arun |
// C# Program to count all possible // groups of size 2 or 3 that have // sum as multiple of 3 using System;
class FindGroups
{ // Returns count of all possible
// groups that can be formed
// from elements of a[].
int findgroups( int []arr, int n)
{
// Create an array C[3] to store
// counts of elements with remainder
// 0, 1 and 2. c[i] would store
// count of elements with remainder i
int [] c= new int []{0, 0, 0};
int i;
// To store the result
int res = 0;
// Count elements with
// remainder 0, 1 and 2
for (i = 0; i < n; i++)
c[arr[i] % 3]++;
// Case 3.a: Count groups of size
// 2 from 0 remainder elements
res += ((c[0] * (c[0] - 1)) >> 1);
// Case 3.b: Count groups of size 2
// with one element with 1 remainder
// and other with 2 remainder
res += c[1] * c[2];
// Case 4.a: Count groups of size 3
// with all 0 remainder elements
res += (c[0] * (c[0] - 1) *
(c[0] - 2)) / 6;
// Case 4.b: Count groups of size 3
// with all 1 remainder elements
res += (c[1] * (c[1] - 1) *
(c[1] - 2)) / 6;
// Case 4.c: Count groups of size 3
// with all 2 remainder elements
res += ((c[2] * (c[2] - 1) *
(c[2] - 2)) / 6);
// Case 4.c: Count groups of size 3
// with different remainders
res += c[0] * c[1] * c[2];
// Return total count stored in res
return res;
}
// Driver Code
public static void Main()
{
FindGroups groups = new FindGroups();
int []arr = {3, 6, 7, 2, 9};
int n = arr.Length;
Console.Write( "Required number of groups are "
+ groups.findgroups(arr, n));
}
} // This code is contributed by nitin mittal. |
<?php // PHP Program to Count groups of size // 2 or 3 that have sum as multiple of 3 // Returns count of all possible groups // that can be formed from elements of a[]. function findgroups( $arr , $n )
{ // Create an array C[3] to store counts
// of elements with remainder 0, 1 and 2.
// c[i] would store count of elements
// with remainder i
$c = array (0, 0, 0);
// To store the result
$res = 0;
// Count elements with remainder
// 0, 1 and 2
for ( $i = 0; $i < $n ; $i ++)
$c [ $arr [ $i ] % 3] += 1;
// Case 3.a: Count groups of size
// 2 from 0 remainder elements
$res += (( $c [0] * ( $c [0] - 1)) >> 1);
// Case 3.b: Count groups of size
// 2 with one element with 1
// remainder and other with 2 remainder
$res += $c [1] * $c [2];
// Case 4.a: Count groups of size
// 3 with all 0 remainder elements
$res += ( $c [0] * ( $c [0] - 1) *
( $c [0] - 2)) / 6;
// Case 4.b: Count groups of size 3
// with all 1 remainder elements
$res += ( $c [1] * ( $c [1] - 1) *
( $c [1] - 2)) / 6;
// Case 4.c: Count groups of size 3
// with all 2 remainder elements
$res += (( $c [2] * ( $c [2] - 1) *
( $c [2] - 2)) / 6);
// Case 4.c: Count groups of size 3
// with different remainders
$res += $c [0] * $c [1] * $c [2];
// Return total count stored in res
return $res ;
} // Driver Code $arr = array (3, 6, 7, 2, 9);
$n = count ( $arr );
echo "Required number of groups are " .
(int)(findgroups( $arr , $n ));
// This code is contributed by mits ?> |
<script> // JavaScript Program to count all possible // groups of size 2 or 3 that have // sum as multiple of 3 // Returns count of all possible groups // that can be formed from elements of a[]. function findgroups(arr, n){
// Create an array C[3] to store counts
// of elements with remainder 0, 1 and 2.
// c[i] would store count of elements
// with remainder i
let c = [0,0,0];
let i;
// To store the result let res = 0;
// Count elements with remainder 0, 1 and 2
for (i=0; i<n; i++)
c[arr[i]%3]++;
// Case 3.a: Count groups of size 2
// from 0 remainder elements
res += ((c[0]*(c[0]-1))>>1);
// Case 3.b: Count groups of size 2 with
// one element with 1 remainder and other
// with 2 remainder
res += c[1] * c[2];
// Case 4.a: Count groups of size 3
// with all 0 remainder elements
res += (c[0] * (c[0]-1) * Math.floor((c[0]-2))/6);
// Case 4.b: Count groups of size 3
// with all 1 remainder elements
res += (c[1] * (c[1]-1) * Math.floor((c[1]-2))/6);
// Case 4.c: Count groups of size 3
// with all 2 remainder elements
res += (Math.floor(c[2]*(c[2]-1)*(c[2]-2))/6);
// Case 4.c: Count groups of size 3
// with different remainders
res += c[0]*c[1]*c[2];
// Return total count stored in res
return res;
} // Driver Code let arr = [3, 6, 7, 2, 9]; let n = arr.length; document.write( "Required number of groups are " + findgroups(arr,n));
</script> |
Required number of groups are 8
Time Complexity: O(n)
Auxiliary Space: O(1)