给定一个由N 个整数组成的数组arr[] ,任务是确定大小为K 的任何子数组中完美数字的最大数量。
例子:
Input: arr[ ] = {28, 2, 3, 6, 496, 99, 8128, 24}, K = 4
Output: 3
Explanation: The sub-array {6, 496, 99, 8128} has 3 perfect numbers which is maximum.
Input: arr[ ]= {1, 2, 3, 6}, K=2
Output: 1
朴素方法:该方法是生成大小为K 的所有可能子数组,并为每个子数组计算完全数元素的数量。打印为任何子数组获得的最大计数。
时间复杂度: O(N*K)
辅助空间: O(1)
有效的方法:
为了优化上述方法,将给定的数组arr[ ]转换为一个二进制数组,如果它是一个完美数,则第i个元素为1 。否则,第 i个元素为0 。因此,问题简化为使用滑动窗口技术在二进制数组中找到大小为K的最大和子数组。请按照以下步骤解决问题:
- 遍历数组,对于数组arr[] 的每个元素,检查它是否是一个完美数。
- 如果arr[i]是完全数,则将arr[i]转换为等于1。否则,将arr[i]转换为等于0 。
- 检查一个数是否为完全数:
- 初始化一个变量sum来存储除数的总和。
- 遍历[1, arr[i] – 1]范围内的每个数字,并检查它是否是arr[i]的除数。添加所有的除数。
- 如果所有除数的总和等于arr[i] ,则该数字是一个完美数字。否则,该数字不是完美数字。
- 计算修改后的数组中第一个大小为K 的子数组的总和。
- 使用滑动窗口技术,从所有可能的大小为K的子数组中找到一个子数组的最大和。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check a number
// is Perfect Number or not
int isPerfect(int N)
{
// Stores sum of divisors
int sum = 1;
// Find all divisors and add them
for (int i = 2; i < sqrt(N); i++)
{
if (N % i == 0) {
if (i == N / i)
{
sum += i;
}
else
{
sum += i + N / i;
}
}
}
// If sum of divisors
// is equal to N
if (sum == N && N != 1)
return 1;
return 0;
}
// Function to return maximum
// sum of a subarray of size K
int maxSum(int arr[], int N, int K)
{
// If k is greater than N
if (N < K)
{
cout << "Invalid";
return -1;
}
// Compute sum of first window of size K
int res = 0;
for (int i = 0; i < K; i++)
{
res += arr[i];
}
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window
int curr_sum = res;
for (int i = K; i < N; i++)
{
curr_sum += arr[i] - arr[i - K];
res = max(res, curr_sum);
}
// return the answer
return res;
}
// Function to find all the
// perfect numbers in the array
int max_PerfectNumbers(int arr[], int N, int K)
{
// The given array is converted into binary array
for (int i = 0; i < N; i++)
{
arr[i] = isPerfect(arr[i]) ? 1 : 0;
}
return maxSum(arr, N, K);
}
// Driver Code
int main()
{
int arr[] = { 28, 2, 3, 6, 496, 99, 8128, 24 };
int K = 4;
int N = sizeof(arr) / sizeof(arr[0]);
cout << max_PerfectNumbers(arr, N, K);
return 0;
}
Java
// Java program for the
// above approach
import java.util.*;
class GFG{
// Function to check a number
// is Perfect Number or not
static int isPerfect(int N)
{
// Stores sum of divisors
int sum = 1;
// Find all divisors and
// add them
for (int i = 2;
i < Math.sqrt(N); i++)
{
if (N % i == 0)
{
if (i == N / i)
{
sum += i;
}
else
{
sum += i + N / i;
}
}
}
// If sum of divisors
// is equal to N
if (sum == N && N != 1)
return 1;
return 0;
}
// Function to return maximum
// sum of a subarray of size K
static int maxSum(int arr[],
int N, int K)
{
// If k is greater than N
if (N < K)
{
System.out.print("Invalid");
return -1;
}
// Compute sum of first
// window of size K
int res = 0;
for (int i = 0; i < K; i++)
{
res += arr[i];
}
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window
int curr_sum = res;
for (int i = K; i < N; i++)
{
curr_sum += arr[i] - arr[i - K];
res = Math.max(res, curr_sum);
}
// return the answer
return res;
}
// Function to find all the
// perfect numbers in the array
static int max_PerfectNumbers(int arr[],
int N, int K)
{
// The given array is converted
// into binary array
for (int i = 0; i < N; i++)
{
arr[i] = isPerfect(arr[i]) ==
1 ? 1 : 0;
}
return maxSum(arr, N, K);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {28, 2, 3, 6, 496,
99, 8128, 24};
int K = 4;
int N = arr.length;
System.out.print(max_PerfectNumbers(arr,
N, K));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program for the above approach
# Function to check a number
# is Perfect Number or not
def isPerfect(N):
# Stores sum of divisors
sum = 1
# Find all divisors and add them
for i in range(2, N):
if i * i > N:
break
if (N % i == 0):
if (i == N // i):
sum += i
else:
sum += i + N // i
# If sum of divisors
# is equal to N
if (sum == N and N != 1):
return 1
return 0
# Function to return maximum
# sum of a subarray of size K
def maxSum(arr, N, K):
# If k is greater than N
if (N < K):
print("Invalid")
return -1
# Compute sum of first
# window of size K
res = 0
for i in range(K):
res += arr[i]
# Compute sums of remaining windows by
# removing first element of previous
# window and adding last element of
# current window
curr_sum = res
for i in range(K, N):
curr_sum += arr[i] - arr[i - K]
res = max(res, curr_sum)
# print(res)
# Return the answer
return res
# Function to find all the
# perfect numbers in the array
def max_PerfectNumbers(arr, N, K):
# The given array is converted
# into binary array
for i in range(N):
if isPerfect(arr[i]):
arr[i] = 1
else:
arr[i] = 0
return maxSum(arr, N, K)
# Driver Code
if __name__ == '__main__':
arr = [ 28, 2, 3, 6,
496, 99, 8128, 24 ]
K = 4
N = len(arr)
print(max_PerfectNumbers(arr, N, K))
# This code is contributed by mohit kumar 29
C#
// C# program for the
// above approach
using System;
class GFG{
// Function to check a number
// is Perfect Number or not
static int isPerfect(int N)
{
// Stores sum of divisors
int sum = 1;
// Find all divisors and
// add them
for (int i = 2;
i < Math.Sqrt(N); i++)
{
if (N % i == 0)
{
if (i == N / i)
{
sum += i;
}
else
{
sum += i + N / i;
}
}
}
// If sum of divisors
// is equal to N
if (sum == N && N != 1)
return 1;
return 0;
}
// Function to return maximum
// sum of a subarray of size K
static int maxSum(int []arr,
int N, int K)
{
// If k is greater than N
if (N < K)
{
Console.Write("Invalid");
return -1;
}
// Compute sum of first
// window of size K
int res = 0;
for (int i = 0; i < K; i++)
{
res += arr[i];
}
// Compute sums of remaining
// windows by removing first
// element of previous window
// and adding last element of
// current window
int curr_sum = res;
for (int i = K; i < N; i++)
{
curr_sum += arr[i] - arr[i - K];
res = Math.Max(res, curr_sum);
}
// return the answer
return res;
}
// Function to find all the
// perfect numbers in the array
static int max_PerfectNumbers(int []arr,
int N, int K)
{
// The given array is converted
// into binary array
for (int i = 0; i < N; i++)
{
arr[i] = isPerfect(arr[i]) ==
1 ? 1 : 0;
}
return maxSum(arr, N, K);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {28, 2, 3, 6, 496,
99, 8128, 24};
int K = 4;
int N = arr.Length;
Console.Write(max_PerfectNumbers(arr,
N, K));
}
}
// This code is contributed by Amit Katiyar
Javascript
输出:
3
时间复杂度: O( N * sqrt(N) )
辅助空间: O(1)
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