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:
Try It Yourself
// 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>
Output
Required number of groups are 8
Time Complexity: O(n)
Auxiliary Space: O(1)