Length of longest subsequence having absolute difference of all pairs divisible by K

Given an array, arr[] of size N and an integer K, the task is to find the length of the longest subsequence from the given array such that the absolute difference of each pair in the subsequence is divisible by K.

Examples:

Input: arr[] = {10, 12, 16, 20, 32, 15}, K = 4  
Output: 4 
Explanation:
The Longest subsequence in which the absolute difference of each pair divisible by K (= 4) are {12, 26, 20, 32}.
Therefore, the required output is 4

Input: arr[] = {12, 3, 13, 5, 21, 11}, K = 3
Output: 3

Naive Approach: The simplest approach to solve this problem is to generate all possible subsequence of the given array and print the length of the longest subsequence having an absolute difference of each pair divisible by K.

Time Complexity: O(2^N)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach the idea is to use Hashing based on the following observation:

Absolute difference of all possible pairs of a subset having the equal value of arr[i] % K must be divisible by K.

Mathematical Proof:
If arr[i] % K = arr[j] % K
=> abs(arr[i] – arr[j]) % K must be 0.

Follow the steps below to solve the problem:

• Initialize an array, say hash[K] to store the frequency of arr[i] % K.
• Traverse the hash[] array and find the maximum element in hash[] array.
• Finally, print the maximum element of the hash[] array.

Below is the implementation of the above approach :

C++14

// C++14 program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the length
// of subsequence that satisfy
// the given condition
int maxLenSub(int arr[],
              int N, int K)
{
    // Store the frequencies
    // of arr[i] % K
    int hash[K];

    // Initialize hash[] array
    memset(hash, 0, sizeof(hash));

    // Traverse the given array
    for (int i = 0; i < N; i++) {

        // Update frequency of
        // arr[i] % K
        hash[arr[i] % K]++;
    }

    // Stores the length of
    // the longest subsequence that
    // satisfy the given condition
    int LenSub = 0;

    // Find the maximum element
    // in hash[] array
    for (int i = 0; i < K; i++) {
        LenSub = max(LenSub, hash[i]);
    }
  return LenSub;
}

// Driver Code
int main()
{
    int arr[] = { 12, 3, 13, 5, 21, 11 };
    int K = 3;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << maxLenSub(arr, N, K);
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{

// Function to find the length
// of subsequence that satisfy
// the given condition
static int maxLenSub(int arr[],
                     int N, int K)
{
  // Store the frequencies
  // of arr[i] % K
  int []hash = new int[K];

  // Traverse the given array
  for (int i = 0; i < N; i++)
  {
    // Update frequency of
    // arr[i] % K
    hash[arr[i] % K]++;
  }

  // Stores the length of
  // the longest subsequence that
  // satisfy the given condition
  int LenSub = 0;

  // Find the maximum element
  // in hash[] array
  for (int i = 0; i < K; i++)
  {
    LenSub = Math.max(LenSub,
                      hash[i]);
  }

  return LenSub;
}

// Driver Code
public static void main(String[] args)
{
  int arr[] = {12, 3, 13, 5, 21, 11};
  int K = 3;
  int N = arr.length;
  System.out.print(maxLenSub(arr, N, K));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above approach

# Function to find the length
# of subsequence that satisfy
# the given condition
def maxLenSub(arr, N, K):

    # Store the frequencies
    # of arr[i] % K
    hash = [0] * K

    # Traverse the given array
    for i in range(N):

        # Update frequency of
        # arr[i] % K
        hash[arr[i] % K] += 1

    # Stores the length of the
    # longest subsequence that
    # satisfy the given condition
    LenSub = 0

    # Find the maximum element
    # in hash[] array
    for i in range(K):
        LenSub = max(LenSub, hash[i])

    return LenSub

# Driver Code
if __name__ == '__main__':

    arr = [ 12, 3, 13, 5, 21, 11 ]
    K = 3
    N = len(arr)

    print(maxLenSub(arr, N, K))

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG{

// Function to find the length
// of subsequence that satisfy
// the given condition
static int maxLenSub(int []arr,
                     int N, int K)
{
  // Store the frequencies
  // of arr[i] % K
  int []hash = new int[K];

  // Traverse the given array
  for (int i = 0; i < N; i++)
  {
    // Update frequency of
    // arr[i] % K
    hash[arr[i] % K]++;
  }

  // Stores the length of
  // the longest subsequence that
  // satisfy the given condition
  int LenSub = 0;

  // Find the maximum element
  // in hash[] array
  for (int i = 0; i < K; i++)
  {
    LenSub = Math.Max(LenSub,
                      hash[i]);
  }

  return LenSub;
}

// Driver Code
public static void Main(String[] args)
{
  int []arr = {12, 3, 13,
               5, 21, 11};
  int K = 3;
  int N = arr.Length;
  Console.Write(maxLenSub(arr, N, K));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to find the length
// of subsequence that satisfy
// the given condition
function maxLenSub(arr, N, K)
{

    // Store the frequencies
    // of arr[i] % K
    let hash = Array.from({length: K}, (_, i) => 0);

    // Traverse the given array
    for(let i = 0; i < N; i++)
    {

        // Update frequency of
        // arr[i] % K
        hash[arr[i] % K]++;
    }

    // Stores the length of
    // the longest subsequence that
    // satisfy the given condition
    let LenSub = 0;

    // Find the maximum element
    // in hash[] array
    for(let i = 0; i < K; i++)
    {
        LenSub = Math.max(LenSub,
                          hash[i]);
    }
    return LenSub;
}

// Driver Code
let arr = [ 12, 3, 13, 5, 21, 11 ];
let K = 3;
let N = arr.length;

document.write(maxLenSub(arr, N, K));

// This code is contributed by target_2

</script>

Output

3

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