# 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:
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(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:

Below is the implementation of the above approach :

 `// C++14 program to implement` `// the above approach` `#include ` `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];`   `    ``// Initilize 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]);` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 12, 3, 13, 5, 21, 11 };` `    ``int` `K = 3;` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << maxLenSub(arr, N, K);` `}`

 `// 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 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# 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`

Output
`3`

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

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