仅具有非负元素的最大和连续子数组

给定一个整数数组arr [] ，任务是找到非负元素的最大和连续子数组并返回其和。
例子：

Input: arr[] = {1, 4, -3, 9, 5, -6} 
Output: 14 
Explanation: 
Subarray [9, 5] is the subarray having maximum sum with all non-negative elements.
Input: arr[] = {12, 0, 10, 3, 11} 
Output: 36 
 

天真的方法：
最简单的方法是在遍历子数组并计算每个有效子数组的和并更新最大和时，生成仅具有非负元素的所有子数组。
时间复杂度： O(N ^ 2)
高效方法：
为了优化上述方法，遍历数组，对于遇到的每个非负元素，继续计算总和。对于遇到的每个负元素，请在与当前总和进行比较之后更新最大总和。将总和重置为0，然后继续下一个元素。
下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to return Largest Sum Contiguous
// Subarray having non-negative number
int maxNonNegativeSubArray(int A[], int N)
{
     
    // Length of given array
    int l = N;
 
    int sum = 0, i = 0;
    int Max = -1;
 
    // Traversing array
    while (i < l)
    {
         
        // Increment i counter to avoid
        // negative elements
        while (i < l && A[i] < 0)
        {
            i++;
            continue;
        }
 
        // Calculating sum of contiguous
        // subarray of non-negative
        // elements
        while (i < l && 0 <= A[i])
        {
            sum += A[i++];
 
            // Update the maximum sum
            Max = max(Max, sum);
        }
 
        // Reset sum
        sum = 0;
    }
 
    // Return the maximum sum
    return Max;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 4, -3, 9, 5, -6 };
     
    int N = sizeof(arr) / sizeof(arr[0]);
     
    cout << maxNonNegativeSubArray(arr, N);
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG {
 
    // Function to return Largest Sum Contiguous
    // Subarray having non-negative number
    static int maxNonNegativeSubArray(int[] A)
    {
        // Length of given array
        int l = A.length;
 
        int sum = 0, i = 0;
 
        int max = -1;
 
        // Traversing array
        while (i < l) {
 
            // Increment i counter to avoid
            // negative elements
            while (i < l && A[i] < 0) {
                i++;
                continue;
            }
 
            // Calculating sum of contiguous
            // subarray of non-negative
            // elements
            while (i < l && 0 <= A[i]) {
 
                sum += A[i++];
 
                // Update the maximum sum
                max = Math.max(max, sum);
            }
 
            // Reset sum
            sum = 0;
        }
 
        // Return the maximum sum
        return max;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int[] arr = { 1, 4, -3, 9, 5, -6 };
 
        System.out.println(maxNonNegativeSubArray(
            arr));
    }
}

Python3

# Python3 program for the above approach
import math
 
# Function to return Largest Sum Contiguous
# Subarray having non-negative number
def maxNonNegativeSubArray(A, N):
     
    # Length of given array
    l = N
     
    sum = 0
    i = 0
    Max = -1
 
    # Traversing array
    while (i < l):
         
        # Increment i counter to avoid
        # negative elements
        while (i < l and A[i] < 0):
            i += 1
            continue
         
        # Calculating sum of contiguous
        # subarray of non-negative
        # elements
        while (i < l and 0 <= A[i]):
            sum += A[i]
            i += 1
             
            # Update the maximum sum
            Max = max(Max, sum)
         
        # Reset sum
        sum = 0;
     
    # Return the maximum sum
    return Max
     
# Driver code
arr = [ 1, 4, -3, 9, 5, -6 ]
 
# Length of array
N = len(arr)
 
print(maxNonNegativeSubArray(arr, N))
 
# This code is contributed by sanjoy_62

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return Largest Sum Contiguous
// Subarray having non-negative number
static int maxNonNegativeSubArray(int[] A)
{
     
    // Length of given array
    int l = A.Length;
 
    int sum = 0, i = 0;
    int max = -1;
 
    // Traversing array
    while (i < l)
    {
         
        // Increment i counter to avoid
        // negative elements
        while (i < l && A[i] < 0)
        {
            i++;
            continue;
        }
 
        // Calculating sum of contiguous
        // subarray of non-negative
        // elements
        while (i < l && 0 <= A[i])
        {
            sum += A[i++];
             
            // Update the maximum sum
            max = Math.Max(max, sum);
        }
 
        // Reset sum
        sum = 0;
    }
 
    // Return the maximum sum
    return max;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 4, -3, 9, 5, -6 };
 
    Console.Write(maxNonNegativeSubArray(arr));
}
}
 
// This code is contributed by chitranayal

输出

14

时间复杂度：O(N)

辅助空间：O(1)

更简单的方法：

Java

public class maxNonNegativeSubArray {
     
    public static void main (String[] args) throws java.lang.Exception{
 
 
        int[] arr = {1, 4, -3, 9, 5, -6};
        int max_so_far=0, max_right_here = 0;
        int start = 0, end = 0, s=0;
 
 
        for(int i=0; i max_so_far){
                max_so_far = max_right_here;
 
                start = s;
                end = i;
            }
        }
 
        System.out.print("Sub Array : ");
        for(int i = start; i <= end; i++){
            System.out.print(arr[i] + " ");
        }
 
        System.out.println();
        System.out.print("Largest Sum : " + max_so_far);
 
    }
}

输出

Sub Array : 9 5 
Largest Sum : 14

时间复杂度：O(n)

空间复杂度：O(1)

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