# Count all possible groups of size 2 or 3 that have sum as multiple of 3

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 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++

`// 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 ` |

*chevron_right*

*filter_none*

## C

`// 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; ` `}` |

*chevron_right*

*filter_none*

## Java

`// 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)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## Python3

`# 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 ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## PHP

`<?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 ` `?> ` |

*chevron_right*

*filter_none*

**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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

## Recommended Posts:

- Maximum difference between groups of size two
- Minimum difference between groups of size two
- Reverse an array in groups of given size
- Reverse a Linked List in groups of given size | Set 1
- Reverse an array in groups of given size | Set 2 (Variations of Set 1 )
- Find the size of largest group where groups are according to the xor of digits
- Reverse an array in groups of given size | Set 3 (Single traversal)
- Maximize number of groups formed with size not smaller than its largest element
- Smallest subarray whose sum is multiple of array size
- Count of largest sized groups while grouping according to product of digits
- Count of elements in an Array whose set bits are in a multiple of K
- Count of integers in an Array whose length is a multiple of K
- Count of elements such that difference between sum of left and right sub arrays is equal to a multiple of k
- Count ways to reach the Nth stair using multiple 1 or 2 steps and a single step 3
- Count Inversions of size three in a given array
- Count distinct elements in every window of size k
- Count the Arithmetic sequences in the Array of size at least 3
- Count of subarrays of size K which is a permutation of numbers from 1 to K
- Count of subarrays of size K with elements having even frequencies
- Count number of increasing subsequences of size k