给定的阵列ARR [],任务是重新排列阵列,以产生最大减小子序列和打印的计 可能的最大子序列数,使得每个数组元素都可以是单个子序列的一部分,并且子序列的长度需要最大化。
例子:
Input: arr[] = {5, 2, 3, 4}
Output: 1
Explanation:
The given array can be rearranged to {5, 4, 3, 2}.
Since every element is occurring only once, the rearranged array forms a decreasing subsequence of maximum possible length.
Input: arr[] = {5, 2, 6, 5, 2, 4, 5, 2}
Output: 3
Explanation:
The given array can be rearranged to {5, 2, 6, 5, 4, 2, 5, 2}.
The 3 decreasing subsequences of maximum possible length possible from the given array are {6, 5, 4, 2}, {5, 2} and {5, 2}
方法:
为了解决这个问题,实际上不需要重新排列给定的数组。由于每个元素都可以是单个子序列的一部分,因此子序列的数量将等于出现频率最高的元素的频率。
请按照以下步骤解决问题:
- 遍历数组,将每个数组元素出现的频率存入一个Hashmap
- 现在,遍历 HashMap 以查找数组中元素的最大频率。
- 打印最大频率作为可能的最大递减子序列的计数。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to count maximum subsequence
void Maximum_subsequence(int A[], int N)
{
// Stores the frequency
// of array elements
unordered_map frequency;
// Stores max frequency
int max_freq = 0;
for (int i = 0; i < N; i++) {
// Update frequency of A[i]
frequency[A[i]]++;
}
for (auto it : frequency) {
// Update max subsequence
if (it.second > max_freq) {
max_freq = it.second;
}
}
// Print the count
cout << max_freq << endl;
}
// Driver Code
int main()
{
int arr[] = { 5, 2, 6, 5, 2, 4, 5, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
Maximum_subsequence(arr, N);
return 0;
}
Java
// Java program for rearrange array
// to generate maximum decreasing
// subsequences
import java.util.HashMap;
import java.util.Map;
public class Main {
// Function to count maximum subsequence
public static void Maximum_subsequence(
int[] A, int N)
{
// Stores the frequency
// of array elements
HashMap frequency
= new HashMap<>();
// Stores maximum frequency
int max_freq = 0;
for (int i = 0; i < N; i++) {
// Update frequency of A[i]
if (frequency.containsKey(A[i])) {
frequency.replace(A[i],
frequency.get(A[i]) + 1);
}
else {
frequency.put(A[i], 1);
}
}
for (Map.Entry it : frequency.entrySet()) {
// Update maximum subsequences
if ((int)it.getValue() > max_freq) {
max_freq = (int)it.getValue();
}
}
// Print the result
System.out.println(max_freq);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 2, 6, 5, 2, 4, 5, 2 };
int N = arr.length;
Maximum_subsequence(arr, N);
}
}
Python3
# Python3 program to implement
# the above approach
# Function to count maximum subsequence
def Maximum_subsequence(A, N):
# Stores the frequency
# of array elements
frequency = dict();
# Stores max frequency
max_freq = 0;
for i in range(N):
if (A[i] in frequency):
frequency[A[i]] += 1
else:
frequency[A[i]] = 1
for it in frequency:
# Update max subsequence
if (frequency[it] > max_freq):
max_freq = frequency[it];
# Print the count
print(max_freq);
# Driver Code
arr = [ 5, 2, 6, 5, 2, 4, 5, 2 ];
Maximum_subsequence(arr, len(arr));
# This code is contributed by grand_master
C#
// C# program for rearrange array
// to generate maximum decreasing
// subsequences
using System;
using System.Collections.Generic;
class GFG{
// Function to count maximum subsequence
public static void Maximum_subsequence(int[] A,
int N)
{
// Stores the frequency
// of array elements
Dictionary frequency = new Dictionary();
// Stores maximum frequency
int max_freq = 0;
for(int i = 0; i < N; i++)
{
// Update frequency of A[i]
if (frequency.ContainsKey(A[i]))
{
frequency[A[i]] = frequency[A[i]] + 1;
}
else
{
frequency.Add(A[i], 1);
}
}
foreach(KeyValuePair it in frequency)
{
// Update maximum subsequences
if ((int)it.Value > max_freq)
{
max_freq = (int)it.Value;
}
}
// Print the result
Console.WriteLine(max_freq);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 5, 2, 6, 5, 2, 4, 5, 2 };
int N = arr.Length;
Maximum_subsequence(arr, N);
}
}
// This code is contributed by Rajput-Ji
Javascript
3
时间复杂度: O(N)
辅助空间: O(N)
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