给定一个数组arr ，任务是找到数组元素的子数组，其总和严格大于其余元素。子数组的大小应该是最小的，总和应该是最大的，并且必须是非递增的。
例子：
Input: arr = [7, 6, 13, 12, 11]
Output: 13 12
Explanation:
The subarray [13, 12] and [13, 11] are minimal such that the sum of their elements is strictly greater than the rest of the elements. However the subarray [13, 12] has the maximum total sum of its elements and hence it is returned in non-increasing order.
Input: arr = [8]
Output: 8
方法：
- 最初我们将对数组arr 进行排序并定义一个名为A的向量。现在我们将首先计算整个数组的总和，然后从后面迭代整个循环，同时连续更新temp 。循环运行直到温度变为零。
- 实际上，从背面迭代循环时发生的情况是，当我们从背面迭代循环时，我们正在考虑最大值。因此，我们可以说我们可以在更短的时间内达到目标条件，也可以通过使用更少的变量。
- 现在，随着循环运行，我们通过将nums[i]添加到它并将nums[i]添加到向量 A来不断更新 temp 的值。最后我们返回代表我们想要的输出结果的向量A。
下面是上述方法的实现：
C++
// C++ program to find the Minimum size Subarray
// with maximum sum in non-increasing order
#include
using namespace std;
// Function to find the Minimum size Subarray
void minSet(vector& nums)
{
vector A;
// sort numbers in descending order
sort(nums.begin(), nums.end());
// get total sum of the given array.
int sum = 0;
for (int i = 0; i < nums.size(); i++)
sum += nums[i];
int temp = 0;
// Loop until the sum of numbers
// is greater than sum/2
for (int i = nums.size() - 1;
i >= 0 && temp <= sum / 2;
i--) {
A.push_back(nums[i]);
temp += nums[i];
}
// Print the Minimum size Subarray
for (int i = 0; i < A.size(); i++)
cout << A[i] << " ";
}
// Driver code
int main()
{
vector vec
= { 7, 6, 13, 13, 12, 11 };
minSet(vec);
return 0;
} Java
// Java program to find the Minimum size Subarray
// with maximum sum in non-increasing order
import java.util.*;
class GFG {
// Function to find the Minimum size Subarray
static void minSet(ArrayList nums) {
ArrayList A = new ArrayList ();
// sort numbers in descending order
Collections.sort(nums);
// get total sum of the given array.
int sum = 0;
for (int i = 0; i= 0 && temp<= sum / 2;
i--) {
A.add(nums.get(i));
temp += nums.get(i);
}
// Print the Minimum size Subarray
for (int i = 0; i gfg = new ArrayList ();
gfg.add(7);
gfg.add(6);
gfg.add(13);
gfg.add(13);
gfg.add(12);
gfg.add(11);
// Function calling
minSet(gfg);
}
}
// This code is contributed by rutvik_56 Python3
# Python3 program to find the Minimum size Subarray
# with maximum sum in non-increasing order
# Function to find the Minimum size Subarray
def minSet(nums) :
A = []
# sort numbers in descending order
nums.sort()
# get total sum of the given array.
sum = 0
for i in range(0,len(nums)):
sum += nums[i]
temp = 0
# Loop until the sum of numbers
# is greater than sum/2
for i in range(len(nums)-1, -1, -1):
if(temp > sum / 2):
break
A.append(nums[i])
temp += nums[i]
# Print the Minimum size Subarray
for i in range(0, len(A)):
print(A[i], end = ' ')
# Driver code
vec = [ 7, 6, 13, 13, 12, 11 ]
minSet(vec);
# This code is contributed by Sanjit_PrasadC#
// C# program to find the Minimum size Subarray
// with maximum sum in non-increasing order
using System;
using System.Collections.Generic;
class GFG {
// Function to find the Minimum size Subarray
static void minSet(List nums) {
List A = new List ();
// sort numbers in descending order
nums.Sort();
// get total sum of the given array.
int sum = 0;
for (int i = 0; i < nums.Count; i++)
sum += nums[i];
int temp = 0;
// Loop until the sum of numbers
// is greater than sum/2
for (int i = nums.Count - 1;
i >= 0 && temp<= sum / 2;
i--) {
A.Add(nums[i]);
temp += nums[i];
}
// Print the Minimum size Subarray
for (int i = 0; i gfg = new List ();
gfg.Add(7);
gfg.Add(6);
gfg.Add(13);
gfg.Add(13);
gfg.Add(12);
gfg.Add(11);
// Function calling
minSet(gfg);
}
}
// This code is contributed by sapnasingh4991 Javascript
输出：
13 13 12时间复杂度： O(N * log N)
辅助空间复杂度： O(N)
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