Count all possible groups of size 2 or 3 that have sum as multiple of 3 | GeeksforGeeks
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.
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.
The idea is to see remainder of every element when divided by 3. A set of elements can form a group only if sun of their remainders is multiple of 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.
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;
}