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 4Input: 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(2N)
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> |
3
Time Complexity: O(N)
Auxiliary Space: O(K)
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