给定一个大小为N的数组arr[] ,任务是找到给定数组中存在的最大和非空子序列。
例子:
Input: arr[] = { 2, 3, 7, 1, 9 }
Output: 22
Explanation:
Sum of the subsequence { arr[0], arr[1], arr[2], arr[3], arr[4] } is equal to 22, which is the maximum possible sum of any subsequence of the array.
Therefore, the required output is 22.
Input: arr[] = { -2, 11, -4, 2, -3, -10 }
Output: 13
Explanation:
Sum of the subsequence { arr[1], arr[3] } is equal to 13, which is the maximum possible sum of any subsequence of the array.
Therefore, the required output is 13.
朴素的方法:解决这个问题最简单的方法是生成数组的所有可能的非空子序列,并计算数组的每个子序列的总和。最后,打印从子序列获得的最大和。
时间复杂度: O(N * 2 N )
辅助空间: O(N)
高效方法:思想是遍历数组,计算数组正元素的总和,打印得到的总和。请按照以下步骤解决问题:
- 检查数组的最大元素是否大于0 。如果发现为真,则遍历数组并打印数组中所有正元素的总和。
- 否则,打印数组中存在的最大元素。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to print the maximum
// non-emepty subsequence sum
int MaxNonEmpSubSeq(int a[], int n)
{
// Stores the maximum non-emepty
// subsequence sum in an array
int sum = 0;
// Stores the largest element
// in the array
int max = *max_element(a, a + n);
if (max <= 0) {
return max;
}
// Traverse the array
for (int i = 0; i < n; i++) {
// If a[i] is greater than 0
if (a[i] > 0) {
// Update sum
sum += a[i];
}
}
return sum;
}
// Driver Code
int main()
{
int arr[] = { -2, 11, -4, 2, -3, -10 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << MaxNonEmpSubSeq(arr, N);
return 0;
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to print the maximum
// non-emepty subsequence sum
static int MaxNonEmpSubSeq(int a[], int n)
{
// Stores the maximum non-emepty
// subsequence sum in an array
int sum = 0;
// Stores the largest element
// in the array
int max = a[0];
for(int i = 1; i < n; i++)
{
if(max < a[i])
{
max = a[i];
}
}
if (max <= 0)
{
return max;
}
// Traverse the array
for (int i = 0; i < n; i++)
{
// If a[i] is greater than 0
if (a[i] > 0)
{
// Update sum
sum += a[i];
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { -2, 11, -4, 2, -3, -10 };
int N = arr.length;
System.out.println(MaxNonEmpSubSeq(arr, N));
}
}
// This code is contributed by divyesh072019
Python3
# Python3 program to implement
# the above approach
# Function to print the maxmimum
# non-emepty subsequence sum
def MaxNonEmpSubSeq(a, n):
# Stores the maxmimum non-emepty
# subsequence sum in an array
sum = 0
# Stores the largest element
# in the array
maxm = max(a)
if (maxm <= 0):
return maxm
# Traverse the array
for i in range(n):
# If a[i] is greater than 0
if (a[i] > 0):
# Update sum
sum += a[i]
return sum
# Driver Code
if __name__ == '__main__':
arr = [ -2, 11, -4, 2, -3, -10 ]
N = len(arr)
print(MaxNonEmpSubSeq(arr, N))
# This code is contributed by mohit kumar 29
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to print the maximum
// non-emepty subsequence sum
static int MaxNonEmpSubSeq(int[] a, int n)
{
// Stores the maximum non-emepty
// subsequence sum in an array
int sum = 0;
// Stores the largest element
// in the array
int max = a[0];
for(int i = 1; i < n; i++)
{
if (max < a[i])
{
max = a[i];
}
}
if (max <= 0)
{
return max;
}
// Traverse the array
for(int i = 0; i < n; i++)
{
// If a[i] is greater than 0
if (a[i] > 0)
{
// Update sum
sum += a[i];
}
}
return sum;
}
// Driver Code
static void Main()
{
int[] arr = { -2, 11, -4, 2, -3, -10 };
int N = arr.Length;
Console.WriteLine(MaxNonEmpSubSeq(arr, N));
}
}
// This code is contributed by divyeshrabadiya07
Javascript
13
时间复杂度: O(N)
辅助空间: O(1)
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