Subarray with no pair sum divisible by K

Given an array of N non-negative integers, task is to find the maximum size of a subarray such that the pairwise sum of the elements of this subarray is not divisible by a given integer, K. Also, print this subarray as well. If there are two or more subarrays that follow the above stated condition, then print the first one from the left. 

Prerequisite : Subset with no pair sum divisible by K

Examples :

Input : arr[] = [3, 7, 1, 9, 2]        
        K = 3
Output : 3
         [3, 7, 1]
3 + 7 = 10, 3 + 1 = 4, 7 + 1 = 8, all are 
not divisible by 3. 
It is not possible to get a subarray of size bigger
than 3 with the above-mentioned property.
[7, 1, 9] is also of the same size but [3, 7, 1] comes first.

Input : arr[] = [2, 4, 4, 3]        
        K = 4
Output : 2
         [2, 4]
2 + 4 = 6 is not divisible by 4. 
It is not possible to get a subarray of size bigger 
than 2 with the above-mentioned property.
[4, 3] is also of the same size but [2, 4] comes first.

Naive Approach: The naive method would be to consider all the subarrays. While considering a subarray, take elements pairwise and compute the sum of the two elements of the pair. If the computed sum is divisible by K, then ignore this subarray and continue with the next subarray. Else, compute the sum of other pairs of this subarray in a similar fashion. If no pair’s sum is a multiple of K, then compare the size of this subarray with the maximum size obtained so far and update if required. 
The time complexity of this method would be O().

Efficient Approach(Using Hashing): We create an empty hash table and insert arr[0] % k into it. Now we traverse remaining elements and maintain a window such that no pair in the window is divisible by k. For every traversed element, we remove starting elements while there exist an element in current window which makes a divisible pair with current element. To check if there is an element in current window, we check if following. 

1. If there is an element x such that (K – x % K) is equal to arr[i] % K 
2. OR arr[i] % k is 0 and it exists in the hash. 

Once we make sure that all elements which can make a pair with arr[i] are removed, we add arr[i] to current window and check if size of current window is more than the maximum window so far. 

Steps to solve this problem:

1. declare a map mp of int key and value.

2. declare variable s=0,e=0,maxs=0,maxe=0.

3. insert mp[arr[0]%k]++.

4. iterate through i=1 till n:

         * declare variable mod=arr[i]%k.

         * while mp[k-mod] is not equal to zero or mod==0 and mp[mod] is not equal to zero:

                      * update mp[arr[s]%k]–.

                      * increment s.

         *update mp[mod]++.

         *increment e.

         *check if ((e-s)>(maxe-maxs)) than update maxe and maxs to e and s.

5. iterating i=maxs till maxe:

         * print arr[i].

Implementation:

C++

// CPP code to find the subarray with
// no pair sum divisible by K
#include<bits/stdc++.h>
using namespace std;

// function to find the subarray with
// no pair sum divisible by k
void subarrayDivisibleByK(int arr[], int n, int k)
{
    // hash table to store the remainders
    // obtained on dividing by K
    map<int,int> mp;

    // s : starting index of the
    // current subarray, e : ending
    // index of the current subarray, maxs :
    // starting index of the maximum
    // size subarray so far, maxe : ending
    // index of the maximum size subarray
    // so far
    int s = 0, e = 0, maxs = 0, maxe = 0;

    // insert the first element in the set
    mp[arr[0] % k]++;

    for (int i = 1; i < n; i++)
    {
        int mod = arr[i] % k;

        // Removing starting elements of current
        // subarray while there is an element in
        // set which makes a pair with mod[i] such
        // that the pair sum is divisible.
        while (mp[k - mod] != 0 ||
              (mod == 0 && mp[mod] != 0))
        {
            mp[arr[s] % k]--;
            s++;
        }

        // include the current element in
        // the current subarray the ending
        // index of the current subarray
        // increments by one
        mp[mod]++;
        e++;

        // compare the size of the current
        // subarray with the maximum size so
        // far
        if ((e - s) > (maxe - maxs))
        {
            maxe = e;
            maxs = s;
        }

    }

    cout << "The maximum size is "
         << maxe - maxs + 1 << " and "
         "the subarray is as follows\n";

    for (int i=maxs; i<=maxe; i++)
        cout << arr[i] << " ";
}

int main()
{
    int k = 3;
    int arr[] = {5, 10, 15, 20, 25};
    int n = sizeof(arr)/sizeof(arr[0]);
    subarrayDivisibleByK(arr, n, k);
    return 0;
}

Java

// Java Program to find the subarray with
// no pair sum divisible by K
import java.io.*;
import java.util.*;

public class GFG {

    // function to find the subarray with
    // no pair sum divisible by k
    static void subarrayDivisibleByK(int []arr,
                                int n, int k)
    {

        // hash table to store the remainders
        // obtained on dividing by K
        int []mp = new int[1000];

        // s : starting index of the
        // current subarray, e : ending
        // index of the current subarray, maxs :
        // starting index of the maximum
        // size subarray so far, maxe : ending
        // index of the maximum size subarray
        // so far
        int s = 0, e = 0, maxs = 0, maxe = 0;

        // insert the first element in the set
        mp[arr[0] % k]++;

        for (int i = 1; i < n; i++)
        {
            int mod = arr[i] % k;

            // Removing starting elements of current
            // subarray while there is an element in
            // set which makes a pair with mod[i] such
            // that the pair sum is divisible.
            while (mp[k - mod] != 0 ||
                (mod == 0 && mp[mod] != 0))
            {
                mp[arr[s] % k]--;
                s++;
            }

            // include the current element in
            // the current subarray the ending
            // index of the current subarray
            // increments by one
            mp[mod]++;
            e++;

            // compare the size of the current
            // subarray with the maximum size so
            // far
            if ((e - s) > (maxe - maxs))
            {
                maxe = e;
                maxs = s;
            }

        }

        System.out.print("The maximum size is "
                            + (maxe - maxs + 1)
        + " and the subarray is as follows\n");

        for (int i = maxs; i <= maxe; i++)
            System.out.print(arr[i] + " ");
    }

    // Driver Code
    public static void main(String args[])
    {
        int k = 3;
        int []arr = {5, 10, 15, 20, 25};
        int n = arr.length;
        subarrayDivisibleByK(arr, n, k);
    }
}

// This code is contributed by
// Manish Shaw (manishshaw1)

Python3

# Python3 Program to find the subarray with
# no pair sum divisible by K

# function to find the subarray with
# no pair sum divisible by k
def subarrayDivisibleByK(arr, n, k) :

    # hash table to store the remainders
    # obtained on dividing by K
    mp = [0] * 1000

    # s : starting index of the
    # current subarray, e : ending
    # index of the current subarray, maxs :
    # starting index of the maximum
    # size subarray so far, maxe : ending
    # index of the maximum size subarray
    # so far
    s = 0; e = 0; maxs = 0; maxe = 0;

    # insert the first element in the set
    mp[arr[0] % k] = mp[arr[0] % k] + 1;

    for i in range(1, n):
        mod = arr[i] % k

        # Removing starting elements of current
        # subarray while there is an element in
        # set which makes a pair with mod[i] such
        # that the pair sum is divisible.
        while (mp[k - mod] != 0 or (mod == 0
                            and mp[mod] != 0)) :
            mp[arr[s] % k] = mp[arr[s] % k] - 1
            s = s + 1

        # include the current element in
        # the current subarray the ending
        # index of the current subarray
        # increments by one
        mp[mod] = mp[mod] + 1
        e = e + 1

        # compare the size of the current
        # subarray with the maximum size so
        # far
        if ((e - s) > (maxe - maxs)) :
            maxe = e
            maxs = s

    print ("The maximum size is {} and the "
                   " subarray is as follows"
                   .format((maxe - maxs + 1)))

    for i in range(maxs, maxe + 1) :
        print ("{} ".format(arr[i]), end="")

# Driver Code
k = 3
arr = [5, 10, 15, 20, 25]
n = len(arr)
subarrayDivisibleByK(arr, n, k)

# This code is contributed by
# Manish Shaw (manishshaw1)

C#

// C# Program to find the subarray with
// no pair sum divisible by K
using System;
using System.Collections;

class GFG {

    // function to find the subarray with
    // no pair sum divisible by k
    static void subarrayDivisibleByK(int []arr,
                                int n, int k)
    {

        // hash table to store the remainders
        // obtained on dividing by K
        int []mp = new int[1000];

        // s : starting index of the
        // current subarray, e : ending
        // index of the current subarray, maxs :
        // starting index of the maximum
        // size subarray so far, maxe : ending
        // index of the maximum size subarray
        // so far
        int s = 0, e = 0, maxs = 0, maxe = 0;

        // insert the first element in the set
        mp[arr[0] % k]++;

        for (int i = 1; i < n; i++)
        {
            int mod = arr[i] % k;

            // Removing starting elements of current
            // subarray while there is an element in
            // set which makes a pair with mod[i] such
            // that the pair sum is divisible.
            while (mp[k - mod] != 0 ||
                (mod == 0 && mp[mod] != 0))
            {
                mp[arr[s] % k]--;
                s++;
            }

            // include the current element in
            // the current subarray the ending
            // index of the current subarray
            // increments by one
            mp[mod]++;
            e++;

            // compare the size of the current
            // subarray with the maximum size so
            // far
            if ((e - s) > (maxe - maxs))
            {
                maxe = e;
                maxs = s;
            }

        }

        Console.Write("The maximum size is " +
                           (maxe - maxs + 1) +
            " and the subarray is as follows\n");

        for (int i = maxs; i <= maxe; i++)
            Console.Write(arr[i] + " ");
    }

    // Driver Code
    public static void Main()
    {
        int k = 3;
        int []arr = {5, 10, 15, 20, 25};
        int n = arr.Length;
        subarrayDivisibleByK(arr, n, k);
    }
}

// This code is contributed by
// Manish Shaw (manishshaw1)

PHP

<?php
// PHP Program to find the
// subarray with no pair
// sum divisible by K

// function to find the subarray
// with no pair sum divisible by k
function subarrayDivisibleByK($arr, $n, $k)
{
    // hash table to store the remainders
    // obtained on dividing by K
    $mp = array_fill(0, 1000, 0);

    // s : starting index of the
    // current subarray, e : ending
    // index of the current subarray, maxs :
    // starting index of the maximum
    // size subarray so far, maxe : ending
    // index of the maximum size subarray
    // so far
    $s = 0;
    $e = 0;
    $maxs = 0;
    $maxe = 0;

    // insert the first
    // element in the set
    $mp[$arr[0] % $k]++;

    for ($i = 1; $i < $n; $i++)
    {
        $mod = $arr[$i] % $k;

        // Removing starting elements
        // of current subarray while
        // there is an element in set
        // which makes a pair with
        // mod[i] such that the pair
        // sum is divisible.
        while ($mp[$k - $mod] != 0 ||
                        ($mod == 0 &&
                         $mp[$mod] != 0))
        {
            $mp[$arr[$s] % $k]--;
            $s++;
        }

        // include the current element in
        // the current subarray the ending
        // index of the current subarray
        // increments by one
        $mp[$mod]++;
        $e++;

        // compare the size of the current
        // subarray with the maximum size so
        // far
        if (($e - $s) > ($maxe - $maxs))
        {
            $maxe = $e;
            $maxs = $s;
        }

    }

    echo ("The maximum size is ".
             ($maxe - $maxs + 1).
          " and the subarray is".
                " as follows\n");

    for ($i = $maxs; $i <= $maxe; $i++)
        echo ($arr[$i]." ");
}

// Driver Code
$k = 3;
$arr = array(5, 10, 15, 20, 25);
$n = count($arr);
subarrayDivisibleByK($arr, $n, $k);

// This code is contributed by
// Manish Shaw (manishshaw1)
?>

Javascript

<script>
      // JavaScript Program to find the subarray with
      // no pair sum divisible by K
      // function to find the subarray with
      // no pair sum divisible by k
      function subarrayDivisibleByK(arr, n, k) {
        // hash table to store the remainders
        // obtained on dividing by K
        var mp = new Array(1000).fill(0);

        // s : starting index of the
        // current subarray, e : ending
        // index of the current subarray, maxs :
        // starting index of the maximum
        // size subarray so far, maxe : ending
        // index of the maximum size subarray
        // so far
        var s = 0,
          e = 0,
          maxs = 0,
          maxe = 0;

        // insert the first element in the set
        mp[arr[0] % k]++;

        for (var i = 1; i < n; i++) {
          var mod = arr[i] % k;

          // Removing starting elements of current
          // subarray while there is an element in
          // set which makes a pair with mod[i] such
          // that the pair sum is divisible.
          while (mp[k - mod] != 0 || (mod == 0 && mp[mod] != 0)) {
            mp[arr[s] % k]--;
            s++;
          }

          // include the current element in
          // the current subarray the ending
          // index of the current subarray
          // increments by one
          mp[mod]++;
          e++;

          // compare the size of the current
          // subarray with the maximum size so
          // far
          if (e - s > maxe - maxs) {
            maxe = e;
            maxs = s;
          }
        }

        document.write(
          "The maximum size is " +
            (maxe - maxs + 1) +
            " and the subarray is as follows<br>"
        );

        for (var i = maxs; i <= maxe; i++) document.write(arr[i] + " ");
      }

      // Driver Code
      var k = 3;
      var arr = [5, 10, 15, 20, 25];
      var n = arr.length;
      subarrayDivisibleByK(arr, n, k);

      // This code is contributed by rdtank.
    </script>

Output

The maximum size is 2 and the subarray is as follows
10 15 

Time Complexity : O(nlogn)
Auxiliary Space: O(n)

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.