最小和连续子数组|套装2

给定一个包含N个整数的数组。任务是找到具有最小(最小)和的连续子数组元素的和。

例子：

Input: arr[] = {3, -4, 2, -3, -1, 7, -5}
Output:-6

Input: arr = {2, 6, 8, 1, 4}
Output: 1

在上一篇文章中已经讨论了一种方法。在这篇文章中，讨论了使用最大和连续子数组方法的解决方案。这是基于以下事实：为了找到最小的连续和，我们可以首先使原始数组的元素为负，即。将每个ar [i]替换为-ar [i]，然后应用Kadane算法。显然，如果这是形成的最大和，则最小和将是该和的负数。

下面是上述方法的实现：

C++

// C++ program for
// Smallest sum contiguous subarray | Set 2
#include 
  
using namespace std;
  
// function to find the smallest sum contiguous subarray
int smallestSumSubarr(int arr[], int n)
{
    // First invert the sign of the elements
    for (int i = 0; i < n; i++)
        arr[i] = -arr[i];
  
    // Apply the normal Kadane algorithm But on the elements
    // Of the array having inverted sign
    int sum_here = arr[0], max_sum = arr[0];
  
    for (int i = 1; i < n; i++) {
  
        sum_here = max(sum_here + arr[i], arr[i]);
        max_sum = max(max_sum, sum_here);
    }
  
    // Invert  the answer to get minimum val
    return (-1) * max_sum;
}
  
// Driver Code
int main()
{
    int arr[] = { 3, -4, 2, -3, -1, 7, -5 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << "Smallest sum: "
         << smallestSumSubarr(arr, n);
    return 0;
}

Java

// Java program for Smallest
// sum contiguous subarray | Set 2
import java.io.*;
  
class GFG 
{
  
// function to find the 
// smallest sum contiguous
// subarray
static int smallestSumSubarr(int arr[],
                             int n)
{
    // First invert the 
    // sign of the elements
    for (int i = 0; i < n; i++)
        arr[i] = -arr[i];
  
    // Apply the normal Kadane 
    // algorithm But on the 
    // elements Of the array 
    // having inverted sign
    int sum_here = arr[0], 
        max_sum = arr[0];
  
    for (int i = 1; i < n; i++) 
    {
        sum_here = Math.max(sum_here + 
                            arr[i], arr[i]);
        max_sum = Math.max(max_sum, 
                           sum_here);
    }
  
    // Invert the answer
    // to get minimum val
    return (-1) * max_sum;
}
  
// Driver Code
public static void main (String[] args) 
{
    int arr[] = {3, -4, 2, -3, 
                -1, 7, -5};
  
    int n = arr.length;
  
    System.out.print("Smallest sum: " + 
            smallestSumSubarr(arr, n));
}
}
  
// This code is contributed 
// by inder_verma.

Python3

# Python3 program for
# Smallest sum contiguous subarray | Set 2
  
# function to find the smallest 
# sum contiguous subarray
def smallestSumSubarr(arr, n):
      
    # First invert the sign of the elements
    for i in range(n):
        arr[i] = -arr[i]
  
    # Apply the normal Kadane algorithm but 
    # on the elements of the array having inverted sign
    sum_here = arr[0]
    max_sum = arr[0]
  
    for i in range(1, n):
  
        sum_here = max(sum_here + arr[i], arr[i])
        max_sum = max(max_sum, sum_here)
  
    # Invert the answer to get minimum val
    return (-1) * max_sum
  
# Driver Code
arr = [3, -4, 2, -3, -1, 7, -5]
  
n = len(arr)
  
print("Smallest sum:", 
       smallestSumSubarr(arr, n))
  
# This code is contributed by Mohit Kumar

C#

// C# program for Smallest
// sum contiguous subarray | Set 2
using System;
class GFG 
{
  
// function to find the 
// smallest sum contiguous
// subarray
static int smallestSumSubarr(int []arr,
                             int n)
{
    // First invert the 
    // sign of the elements
    for (int i = 0; i < n; i++)
        arr[i] = -arr[i];
  
    // Apply the normal Kadane 
    // algorithm But on the 
    // elements Of the array 
    // having inverted sign
    int sum_here = arr[0], 
        max_sum = arr[0];
  
    for (int i = 1; i < n; i++) 
    {
        sum_here = Math.Max(sum_here + 
                            arr[i], arr[i]);
        max_sum = Math.Max(max_sum, 
                           sum_here);
    }
  
    // Invert the answer
    // to get minimum val
    return (-1) * max_sum;
}
  
// Driver Code
public static void Main () 
{
    int []arr = {3, -4, 2, -3, 
                -1, 7, -5};
  
    int n = arr.Length;
  
    Console.WriteLine("Smallest sum: " + 
             smallestSumSubarr(arr, n));
}
}
  
// This code is contributed 
// by inder_verma.

输出：

Smallest sum: -6

时间复杂度： O(n)