# 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 â€˜nâ€™, 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.

**Implementation:**

## 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` |

## 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;` `}` |

## 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));` ` ` `}` `}` |

## 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` |

## 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.` |

## 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` `?>` |

## Javascript

`<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)