给定一个大小为N的数组arr[]和一个整数K ,任务是从给定数组中找到最长子序列的长度,使得子序列中每对的绝对差可以被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
朴素方法: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,并打印最长子序列的长度,该长度具有每对可被K整除的绝对差。
时间复杂度: O(2 N )
辅助空间: O(N)
高效的方法:为了优化上述方法,想法是基于以下观察使用哈希:
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,比如hash[K]来存储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];
// 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]);
}
}
// 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
3
时间复杂度: O(N)
辅助空间: O(K)
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