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 groupsint 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 Codeint 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 approachimport 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 groupsdef 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 codeif __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 approachusing 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>
|
Output
4
Time Complexity: O(N)
Auxiliary Space: O(K)
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