具有可被K整除的所有对的绝对差的最长子序列的长度

给定大小为N且整数K的数组arr [] ，任务是从给定数组中找到最长子序列的长度，以使子序列中每一对的绝对差可被K整除。

例子：

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

天真的方法：解决此问题的最简单方法是生成给定数组的所有可能子序列，并打印最长的子序列的长度，每个子序列的绝对差可被K整除。

时间复杂度： O(2 ^N )
辅助空间： O(N)

高效方法：为了优化上述方法，其思想是基于以下观察结果使用哈希：

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.

请按照以下步骤解决问题：

• 初始化一个数组，用hash [K]表示，以存储arr [i]％K的频率。
• 遍历散列[]数组并找到散列[]数组的最大元素。
• 最后，打印hash []数组的最大元素。

下面是上述方法的实现：

C++14

// 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[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

输出

3

时间复杂度： O(N)
辅助空间： O(K)