Minimum count of groups such that sum of adjacent elements is divisible by K in each group
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:
- Groups should consist of only one element which is not divisible by K.
- All elements of the groups should be individually divisible by K.
- 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 <bits/stdc++.h> 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[0]); 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<Integer, Integer> 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
<script> // JavaScript program for the above approach // Function to count the minimum number // of groups function findMinGroups(arr, N, K) { // Stores the count of elements let mp = new Map(); // Stores the count of groups let ans = 0; // Traverse the array arr[] for (let 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.has(arr[i])) { mp.set(arr[i], 1); } else { let ct = mp.get(arr[i]); if (ct != null ) { ct++; mp.set(arr[i], ct); } } } } // Iterarte over the range [1, K / 2] for (let i = 1; i <= K / 2; i++) { // Stores the minimum of count of i // and K - i let a = 0, b = 0; if (mp.has(K - i)) { a = mp.get(K - i); } if (mp.has(i)) { b = mp.get(i); } let c1 = Math.min(a, b); // Stores the maximum of count of i // and K - i let 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; } // Input let arr = [1, 1, 4, 4, 8, 6, 7]; let N = 7; let K = 8; // Function Call document.write(findMinGroups(arr, N, K)); // This code is contributed by Potta Lokesh </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(K)
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