给定一个 array arr [] ，任务是打印所有其元素乘积小于或等于K的可能子数组。
Input: arr[] = {2, 1, 3, 4, 5, 6, 2}, K = 10
Output: [[2], [1], [2, 1], [3], [1, 3], [2, 1, 3], [4], [5], [6], [2]]
Explanation:
All possible subarrays having product ≤ K are {2}, {1}, {2, 1}, {3}, {1, 3}, {2, 1, 3}, {4}, {5}, {6}, {2}.
Input: arr[] = {2, 7, 1, 4}, K = 7
Output: [[2], [7], [1], [7, 1], [4], [1, 4]]
天真的方法：解决问题的最简单方法是从给定数组生成所有可能的子数组，并针对每个子数组检查其乘积是否小于或等于K并进行相应打印。
时间复杂度： O(N 3 )
辅助空间： O(N)
高效的方法：可以通过观察以下方法来优化上述方法：
If the product of all the elements of a subarray is less than or equal to K, then all the subarrays possible from this subarray also has product less than or equal to K. Therefore, these subarrays need to be included in the answer as well.
请按照以下步骤解决问题：
- 初始化一个指针，开始指向数组的第一个索引。
- 遍历数组，并继续计算数组元素的乘积，并将其存储在变量中，例如multi 。
- 如果多超过K：保持除以ARR多[开始]和不断递增开始直到多降至≤ķ。
- 如果multi≤K ：从当前索引迭代到start ，并将子数组存储在Arraylist中。
- 最后，一旦生成了所有子数组，请打印包含所有获得的子数组的Arraylist 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to return all possible
// subarrays having product less
// than or equal to K
vector> maxSubArray(int arr[], int n,
int K)
{
// Store the required subarrays
vector> solution;
// Stores the product of
// current subarray
int multi = 1;
// Stores the starting index
// of the current subarray
int start = 0;
// Check for empty array
if (n <= 1 || K < 0)
{
return solution;
}
// Iterate over the array
for(int i = 0; i < n; i++)
{
// Calculate product
multi = multi * arr[i];
// If product exceeds K
while (multi > K)
{
// Reduce product
multi = multi / arr[start];
// Increase starting index
// of current subarray
start++;
}
// Stores the subarray elements
vector list;
// Store the subarray elements
for(int j = i; j >= start; j--)
{
list.insert(list.begin(), arr[j]);
// Add the subarrays
// to the list
solution.push_back(list);
}
}
// Return the final
// list of subarrays
return solution;
}
// Driver Code
int main()
{
int arr[] = { 2, 7, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 7;
vector> v = maxSubArray(arr, n, K);
cout << "[";
bool first = true;
for(auto x : v)
{
if (!first)
{
cout << ", ";
}
else
{
first = false;
}
cout << "[";
bool ff = true;
for(int y : x)
{
if (!ff)
{
cout << ", ";
}
else
{
ff = false;
}
cout << y;
}
cout << "]";
}
cout << "]";
return 0;
}
// This code is contributed by rutvik_56 Java
// Java Program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to return all possible
// subarrays having product less
// than or equal to K
public static List > maxSubArray(
int[] arr, int K)
{
// Store the required subarrays
List > solution
= new ArrayList<>();
// Stores the product of
// current subarray
int multi = 1;
// Stores the starting index
// of the current subarray
int start = 0;
// Check for empty array
if (arr.length <= 1 || K < 0) {
return new ArrayList<>();
}
// Iterate over the array
for (int i = 0; i < arr.length; i++) {
// Calculate product
multi = multi * arr[i];
// If product exceeds K
while (multi > K) {
// Reduce product
multi = multi / arr[start];
// Increase starting index
// of current subarray
start++;
}
// Stores the subarray elements
List list
= new ArrayList<>();
// Store the subarray elements
for (int j = i; j >= start; j--) {
list.add(0, arr[j]);
// Add the subarrays
// to the list
solution.add(
new ArrayList<>(list));
}
}
// Return the final
// list of subarrays
return solution;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 2, 7, 1, 4 };
int K = 7;
System.out.println(maxSubArray(arr, K));
}
} Python3
# Python3 program to implement
# the above approach
# Function to return all possible
# subarrays having product less
# than or equal to K
def maxSubArray(arr, n, K):
# Store the required subarrays
solution = []
# Stores the product of
# current subarray
multi = 1
# Stores the starting index
# of the current subarray
start = 0
# Check for empty array
if (n <= 1 or K < 0):
return solution
# Iterate over the array
for i in range(n):
# Calculate product
multi = multi * arr[i]
# If product exceeds K
while (multi > K):
# Reduce product
multi = multi // arr[start]
# Increase starting index
# of current subarray
start += 1
# Stores the subarray elements
li = []
j = i
# Store the subarray elements
while(j >= start):
li.insert(0, arr[j])
# Add the subarrays
# to the li
solution.append(list(li))
j -= 1
# Return the final
# li of subarrays
return solution
# Driver Code
if __name__=='__main__':
arr = [ 2, 7, 1, 4 ]
n = len(arr)
K = 7
v = maxSubArray(arr, n, K)
print(v)
# This code is contributed by pratham76C#
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return all possible
// subarrays having product less
// than or equal to K
public static List> maxSubArray(int[] arr,
int K)
{
// Store the required subarrays
List > solution = new List>();
// Stores the product of
// current subarray
int multi = 1;
// Stores the starting index
// of the current subarray
int start = 0;
// Check for empty array
if (arr.Length <= 1 || K < 0)
{
return new List>();
}
// Iterate over the array
for (int i = 0; i < arr.Length; i++)
{
// Calculate product
multi = multi * arr[i];
// If product exceeds K
while (multi > K)
{
// Reduce product
multi = multi / arr[start];
// Increase starting index
// of current subarray
start++;
}
// Stores the subarray elements
List list = new List();
// Store the subarray elements
for (int j = i; j >= start; j--)
{
list.Insert(0, arr[j]);
// Add the subarrays
// to the list
solution.Add(new List(list));
}
}
// Return the final
// list of subarrays
return solution;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = {2, 7, 1, 4};
int K = 7;
List > list = maxSubArray(arr, K);
foreach(List i in list)
{
Console.Write("[");
foreach(int j in i)
{
Console.Write(j);
if(i.Count > 1)
Console.Write(",");
}
Console.Write("]");
}
}
}
// This code is contributed by 29AjayKumar 输出：
[[2], [7], [1], [7, 1], [4], [1, 4]]时间复杂度： O(N 2 )
辅助空间： O(1)