给定一个数组arr []和一个整数K ,任务是计算可以选择的最大数组元素数,以便:
- 所选数组元素的总和小于K。
- K –(选定元素的总和)大于或等于0,并且可能的最小值。
例子:
Examples:
Input: arr[] = {20, 90, 40, 90}, K = 100
Output: 2
Explanation: {20, 40} are selected such that their sum ( = 60) is less than K and K – total sum ( = 40) is greater than 0 and minimum possible.
Examples:
Input: arr[] = {30, 30, 10, 10}, K = 50
Output: 3
Explanation: {10, 10, 30} are selected< such that their sum ( = 50) is equal to K and K – total sum is 0 and minimum possible.
方法:解决此问题的主要思想是使用贪婪方法。请按照以下步骤解决此问题:
- 对给定的数组进行排序。
- 现在,从第一个数组元素开始,将当前元素添加到总和中。
- 增量计数为1。
- 检查总和是否大于K。
- 如果发现是真实的,则不要选择任何其他元素。
- 否则,重复上述过程,直到遍历数组的最后一个元素。
- 答案是所选元素的总数。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to count maximum number
// of elements that can be selected
int CountMaximum(int arr[], int n, int k)
{
// Sort he array
sort(arr, arr + n);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add current element
// to the sum
sum += arr[i];
// IF sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Return the count
return count;
}
// Driver Code
int main()
{
int arr[] = { 30, 30, 10, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 50;
cout << CountMaximum(arr, n, k);
return 0;
}
Java
// Java implementation of the above approach
import java.util.Arrays;
public class GFG {
// Function to count maximum number
// of elements that can be selected
static int CountMaximum(int arr[], int n, int k)
{
// Sorting the array
Arrays.sort(arr);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add the current
// element to the sum
sum += arr[i];
// If sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Returning the count
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 30, 30, 10, 10 };
int n = 4;
int k = 50;
// Function call
System.out.println(
CountMaximum(arr, n, k));
}
}
Python3
# Python3 implementation of the above approach
# Function to count maximum number
# of elements that can be selected
def CountMaximum(arr, n, k) :
# Sort he array
arr.sort()
Sum, count = 0, 0
# Traverse the array
for i in range(0, n) :
# Add current element
# to the sum
Sum += arr[i]
# IF sum exceeds k
if (Sum > k) :
break
# Increment count
count += 1
# Return the count
return count
# Driver code
arr = [ 30, 30, 10, 10 ]
n = len(arr)
k = 50
print(CountMaximum(arr, n, k))
# This code is contributed by divyesh072019
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to count maximum number
// of elements that can be selected
static int CountMaximum(int[] arr, int n, int k)
{
// Sorting the array
Array.Sort(arr);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Add the current
// element to the sum
sum += arr[i];
// If sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Returning the count
return count;
}
// Driver code
static public void Main()
{
int[] arr = new int[] { 30, 30, 10, 10 };
int n = 4;
int k = 50;
// Function call
Console.WriteLine(CountMaximum(arr, n, k));
}
}
// This code is contributed by dharanendralv23
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to count maximum number
// of elements that can be selected
int CountMaximum(int arr[], int n, int k)
{
// Sort the array
sort(arr, arr + n);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add current element
// to the sum
sum += arr[i];
// IF sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Return the count
return count;
}
// Driver Code
int main()
{
int arr[] = { 30, 30, 10, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 50;
cout << CountMaximum(arr, n, k);
return 0;
}
Java
// Java implementation of the above approach
import java.util.Arrays;
public class GFG {
// Function to count maximum number
// of elements that can be selected
static int CountMaximum(int arr[], int n, int k)
{
// Sorting the array
Arrays.sort(arr);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add the current
// element to the sum
sum += arr[i];
// If sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Returning the count
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 30, 30, 10, 10 };
int n = 4;
int k = 50;
// Function call
System.out.println(
CountMaximum(arr, n, k));
}
}
Python3
# Python3 implementation of the above approach
# Function to count maximum number
# of elements that can be selected
def CountMaximum(arr, n, k) :
# Sort he array
arr.sort()
Sum, count = 0, 0
# Traverse the array
for i in range(0, n) :
# Add current element
# to the sum
Sum += arr[i]
# IF sum exceeds k
if (Sum > k) :
break
# Increment count
count += 1
# Return the count
return count
# Driver code
arr = [ 30, 30, 10, 10 ]
n = len(arr)
k = 50
print(CountMaximum(arr, n, k))
# This code is contributed by divyesh072019
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to count maximum number
// of elements that can be selected
static int CountMaximum(int[] arr, int n, int k)
{
// Sorting the array
Array.Sort(arr);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Add the current
// element to the sum
sum += arr[i];
// If sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Returning the count
return count;
}
// Driver code
static public void Main()
{
int[] arr = new int[] { 30, 30, 10, 10 };
int n = 4;
int k = 50;
// Function call
Console.WriteLine(CountMaximum(arr, n, k));
}
}
// This code is contributed by dharanendralv23
Javascript
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to count maximum number
// of elements that can be selected
int CountMaximum(int arr[], int n, int k)
{
// Sort the array
sort(arr, arr + n);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add current element
// to the sum
sum += arr[i];
// IF sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Return the count
return count;
}
// Driver Code
int main()
{
int arr[] = { 30, 30, 10, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 50;
cout << CountMaximum(arr, n, k);
return 0;
}
Java
// Java implementation of the above approach
import java.util.Arrays;
public class GFG {
// Function to count maximum number
// of elements that can be selected
static int CountMaximum(int arr[], int n, int k)
{
// Sorting the array
Arrays.sort(arr);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Add the current
// element to the sum
sum += arr[i];
// If sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Returning the count
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 30, 30, 10, 10 };
int n = 4;
int k = 50;
// Function call
System.out.println(
CountMaximum(arr, n, k));
}
}
Python3
# Python3 implementation of the above approach
# Function to count maximum number
# of elements that can be selected
def CountMaximum(arr, n, k) :
# Sort he array
arr.sort()
Sum, count = 0, 0
# Traverse the array
for i in range(0, n) :
# Add current element
# to the sum
Sum += arr[i]
# IF sum exceeds k
if (Sum > k) :
break
# Increment count
count += 1
# Return the count
return count
# Driver code
arr = [ 30, 30, 10, 10 ]
n = len(arr)
k = 50
print(CountMaximum(arr, n, k))
# This code is contributed by divyesh072019
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to count maximum number
// of elements that can be selected
static int CountMaximum(int[] arr, int n, int k)
{
// Sorting the array
Array.Sort(arr);
int sum = 0, count = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Add the current
// element to the sum
sum += arr[i];
// If sum exceeds k
if (sum > k)
break;
// Increment count
count++;
}
// Returning the count
return count;
}
// Driver code
static public void Main()
{
int[] arr = new int[] { 30, 30, 10, 10 };
int n = 4;
int k = 50;
// Function call
Console.WriteLine(CountMaximum(arr, n, k));
}
}
// This code is contributed by dharanendralv23
Java脚本
输出:
3
时间复杂度: O(N log N)
辅助空间: O(1)