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.

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 sun 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.

## C

`#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

`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

`# Python 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#

`// 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. ` |

Output:

Required number of groups are 8

Time Complexity: O(n)

Auxiliary Space: O(1)

This article is contributed by Amit Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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