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# Minimum count of groups such that sum of adjacent elements is divisible by K in each group

• Difficulty Level : Medium
• Last Updated : 28 Jul, 2021

Given an array arr[] consisting of N integers, and an integer K, the task is to print the minimum count of groups formed by taking elements of the array such that the sum of adjacent elements is divisible by K in each group.

Examples:

Input: arr[] = {2, 6, 5, 8, 2, 9}, K = 2
Output: 2
The array can be split into two groups {2, 6, 2, 8} and {5, 9}.
The sum of adjacent elements of each group is divisible by K.
Thus, the minimum number of groups that can be formed is 2.

Input: arr[] = {1, 1, 4, 4, 8, 6, 7}, K = 8
Output: 4
Explanation:
The array can be split into 4 groups: {1, 7, 1}, {4, 4}, {8}, and {6}.
The sum of adjacent elements of each group is divisible by K.
Thus, the minimum number of groups that can be formed is 4.

Input: arr[] = {144}, K = 5
Output: 1

Approach: The given problem can be solved based on the following observations:

• It can be observed that a group should be formed as:
1. Groups should consist of only one element which is not divisible by K.
2. All elements of the groups should be individually divisible by K.
3. Every adjacent element of the group should be satisfied; X % K + Y % K = K, where X and Y are two adjacent elements of the group.

Follow the steps below to solve the problem:

• Initialize a map<int, int>, say mp, to store the count of arr[i] % K.
• Initialize a variable, say ans as 0, to store the count of groups.
• Traverse the array arr[] and increment the count of arr[i] % K in the mp and if arr[i] % K is 0 then assign 1 to ans.
• Iterate over the range [1, K / 2] and perform the following operations:
• Store the minimum of mp[i] and mp[K – i] in a variable, say C1.
• Store the maximum of mp[i] and mp[K – i] in a variable, say C2.
• If C1 is 0 then increment ans by C2.
• Otherwise, if C1 is either equal to C1 or C1 + 1, then increment ans by 1.
• Otherwise, increment ans by C2 – C1 – 1.
• Finally, after completing the above steps, print the answer obtained in ans.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to count the minimum number``// of groups``int` `findMinGroups(``int` `arr[], ``int` `N, ``int` `K)``{``    ``// Stores the count of elements``    ``unordered_map<``int``, ``int``> mp;` `    ``// Stores the count of groups``    ``int` `ans = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Update arr[i]``        ``arr[i] = arr[i] % K;` `        ``// If arr[i] is 0``        ``if` `(arr[i] == 0) {` `            ``// Update ans``            ``ans = 1;``        ``}``        ``else` `{` `            ``// Increment mp[arr[i]] by 1``            ``mp[arr[i]]++;``        ``}``    ``}` `    ``// Iterarte over the range [1, K / 2]``    ``for` `(``int` `i = 1; i <= K / 2; i++) {` `        ``// Stores the minimum of count of i``        ``// and K - i``        ``int` `c1 = min(mp[K - i], mp[i]);` `        ``// Stores the maximum of count of i``        ``// and K - i``        ``int` `c2 = max(mp[K - i], mp[i]);` `        ``// If c1 is 0``        ``if` `(c1 == 0) {` `            ``// Increment ans by c2``            ``ans += c2;``        ``}` `        ``// Otherwise if c2 is equal to c1 + 1``        ``// or c1``        ``else` `if` `(c2 == c1 + 1 || c1 == c2) {` `            ``// Increment ans by 1``            ``ans++;``        ``}``        ``// Otherwise``        ``else` `{` `            ``// Increment ans by c2 - c1 - 1``            ``ans += (c2 - c1 - 1);``        ``}``    ``}` `    ``// Return the ans``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``// Input``    ``int` `arr[] = { 1, 1, 4, 4, 8, 6, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `K = 8;` `    ``// Function Call``    ``cout << findMinGroups(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{``  ` `    ``// Function to count the minimum number``    ``// of groups``    ``public` `static` `int` `findMinGroups(``int` `arr[], ``int` `N, ``int` `K)``    ``{``        ``// Stores the count of elements``        ``HashMap mp = ``new` `HashMap<>();``        ` `        ``// Stores the count of groups``        ``int` `ans = ``0``;``    ` `        ``// Traverse the array arr[]``        ``for` `(``int` `i = ``0``; i < N; i++) {``    ` `            ``// Update arr[i]``            ``arr[i] = arr[i] % K;``    ` `            ``// If arr[i] is 0``            ``if` `(arr[i] == ``0``) {``    ` `                ``// Update ans``                ``ans = ``1``;``            ``}``            ``else` `{``    ` `                ``// Increment mp[arr[i]] by 1``                ``if` `(!mp.containsKey(arr[i])) {``                    ``mp.put(arr[i], ``1``);``                ``}``                ``else` `{``                    ``Integer ct = mp.get(arr[i]);``                    ``if``(ct!=``null``)``                    ``{``                        ``ct++;``                        ``mp.put(arr[i], ct);``                    ``}``                ``}``            ``}``        ``}``    ` `        ``// Iterarte over the range [1, K / 2]``        ``for` `(``int` `i = ``1``; i <= K / ``2``; i++) {``    ` `            ``// Stores the minimum of count of i``            ``// and K - i``            ``int` `a=``0``,b=``0``;``            ``if``(mp.containsKey(K-i)){``                ``a=mp.get(K-i);``            ``}``            ``if``(mp.containsKey(i)){``                ``b=mp.get(i);``            ``}``            ``int` `c1 = Math.min(a, b);``    ` `            ``// Stores the maximum of count of i``            ``// and K - i``            ``int` `c2 = Math.max(a, b);``    ` `            ``// If c1 is 0``            ``if` `(c1 == ``0``) {``    ` `                ``// Increment ans by c2``                ``ans += c2;``            ``}``    ` `            ``// Otherwise if c2 is equal to c1 + 1``            ``// or c1``            ``else` `if` `(c2 == c1 + ``1` `|| c1 == c2) {``    ` `                ``// Increment ans by 1``                ``ans++;``            ``}``            ``// Otherwise``            ``else` `{``    ` `                ``// Increment ans by c2 - c1 - 1``                ``ans += (c2 - c1 - ``1``);``            ``}``        ``}``    ` `        ``// Return the ans``        ``return` `ans;``    ``}``    ``public` `static` `void` `main (String[] args) {``        ``// Input``        ``int` `arr[] = { ``1``, ``1``, ``4``, ``4``, ``8``, ``6``, ``7` `};``        ``int` `N = ``7``;``        ``int` `K = ``8``;``        ` `        ``// Function Call``        ``System.out.println(findMinGroups(arr, N, K));``    ``}``}` `// This code is contributed by Manu Pathria`

## Python3

 `# Python3 program for the above approach` `# Function to count the minimum number``# of groups``def` `findMinGroups(arr, N, K):``    ` `    ``# Stores the count of elements``    ``mp ``=` `{}` `    ``# Stores the count of groups``    ``ans ``=` `0` `    ``# Traverse the array arr[]``    ``for` `i ``in` `range``(N):``        ` `        ``# Update arr[i]``        ``arr[i] ``=` `arr[i] ``%` `K` `        ``# If arr[i] is 0``        ``if` `(arr[i] ``=``=` `0``):``            ` `            ``# Update ans``            ``ans ``=` `1``        ``else``:``            ``mp[arr[i]] ``=` `mp.get(arr[i], ``0``) ``+` `1` `    ``# Iterarte over the range [1, K / 2]``    ``for` `i ``in` `range``(``1``, K ``/``/` `2``):``        ` `        ``# Stores the minimum of count of i``        ``# and K - i``        ``x, y ``=` `(``0` `if` `(K ``-` `i) ``not` `in` `mp ``else` `mp[K ``-` `i],``                      ``0` `if` `i ``not` `in` `mp ``else` `mp[K ``-` `i])``        ``c1 ``=` `min``(x, y)` `        ``# The maximum of count of i``        ``# K - i``        ``c2 ``=` `max``(x, y)` `        ``# If c1 is 0``        ``if` `(c1 ``=``=` `0``):``            ` `            ``# Increment ans by c2``            ``ans ``+``=` `c2``            ` `        ``# Otherwise if c2 is equal to c1 + 1``        ``# or c1``        ``elif` `(c2 ``=``=` `c1 ``+` `1` `or` `c1 ``=``=` `c2):``            ` `            ``# Increment ans by 1``            ``ans ``+``=` `1``            ` `        ``# Otherwise   ``        ``else``:` `            ``# Increment ans by c2 - c1 - 1``            ``ans ``+``=` `(c2 ``-` `c1 ``-` `1``)``            ` `    ``# Return the ans``    ``return` `ans ``+` `1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Input``    ``arr ``=` `[ ``1``, ``1``, ``4``, ``4``, ``8``, ``6``, ``7` `]``    ``N ``=` `len``(arr)``    ``K ``=` `8` `    ``# Function Call``    ``print``(findMinGroups(arr, N, K))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG{` `    ``// Function to count the minimum number``    ``// of groups``     ``static` `int` `findMinGroups(``int``[] arr, ``int` `N, ``int` `K)``    ``{``        ``// Stores the count of elements``        ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``, ``int``>();``        ` `        ``// Stores the count of groups``        ``int` `ans = 0;``    ` `        ``// Traverse the array arr[]``        ``for` `(``int` `i = 0; i < N; i++) {``    ` `            ``// Update arr[i]``            ``arr[i] = arr[i] % K;``    ` `            ``// If arr[i] is 0``            ``if` `(arr[i] == 0) {``    ` `                ``// Update ans``                ``ans = 1;``            ``}``            ``else` `{``    ` `                ``// Increment mp[arr[i]] by 1``                ``if` `(!mp.ContainsKey(arr[i])) {``                    ``mp.Add(arr[i], 1);``                ``}``                ``else` `{``                    ``int` `ct = mp[arr[i]];``                    ``if``(ct!=0)``                    ``{``                        ``ct++;``                        ``mp[arr[i]]= ct;``                    ``}``                ``}``            ``}``        ``}``    ` `        ``// Iterarte over the range [1, K / 2]``        ``for` `(``int` `i = 1; i <= K / 2; i++) {``    ` `            ``// Stores the minimum of count of i``            ``// and K - i``            ``int` `a=0,b=0;``            ``if``(mp.ContainsKey(K-i)){``                ``a=mp[K-i];``            ``}``            ``if``(mp.ContainsKey(i)){``                ``b=mp[i];``            ``}``            ``int` `c1 = Math.Min(a, b);``    ` `            ``// Stores the maximum of count of i``            ``// and K - i``            ``int` `c2 = Math.Max(a, b);``    ` `            ``// If c1 is 0``            ``if` `(c1 == 0) {``    ` `                ``// Increment ans by c2``                ``ans += c2;``            ``}``    ` `            ``// Otherwise if c2 is equal to c1 + 1``            ``// or c1``            ``else` `if` `(c2 == c1 + 1 || c1 == c2) {``    ` `                ``// Increment ans by 1``                ``ans++;``            ``}``            ``// Otherwise``            ``else` `{``    ` `                ``// Increment ans by c2 - c1 - 1``                ``ans += (c2 - c1 - 1);``            ``}``        ``}``    ` `        ``// Return the ans``        ``return` `ans;``    ``}``    ``static` `public` `void` `Main ()``    ``{``        ``// InAdd``        ``int``[] arr = { 1, 1, 4, 4, 8, 6, 7 };``        ``int` `N = 7;``        ``int` `K = 8;``        ` `        ``// Function Call``        ``Console.Write(findMinGroups(arr, N, K));``    ``}``}` `// This code is contributed by shubhamsingh10`

## Javascript

 ``
Output
`4`

Time Complexity: O(N)
Auxiliary Space: O(K)

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