要添加到每个数组索引的前缀总和的最小值以使它们为正

给定一个由N个整数组成的数组 arr [] ，任务是找到需要添加的最小正值S ，使得添加S后给定数组的每个索引处的前缀和始终为正。

例子：

Input: arr[] = {-3, 2, -3, 4, 2}
Output: 5
Explanation:
For S = 5, prefix sums at each array indices are as follows:
At index 1: (5 – 3) = 2
At index 2: (2 + 2) = 4
At index 3: (4 – 3) = 1
At index 4: (1 + 4) = 5
At index 5: (5 + 2) = 7
Since all the prefix sums obtained is greater than 0, the required answer is 5.

Input: arr[] = {1, 2}
Output: 1

朴素的方法：最简单的方法是从1开始迭代S 的所有可能值，并将S添加到每个数组索引的前缀和，并检查它是否大于0 。打印所有前缀和都为正的S值。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：要优化上述方法，请按照以下步骤操作：

• 用0初始化变量minValue以存储最小前缀和。
• 初始化一个变量sum以在每个索引处存储前缀和。
• 使用变量i在范围[0, N – 1] 上遍历数组arr[] 。
  • 更新sum并将arr[i]添加到sum 。
  • 如果sum小于minValue ，则将minValue设置为sum 。
• 初始化变量startValue以存储所需的结果并将startValue设置为等于(1 – minValue) 。
• 完成上述步骤后，将startValue的值打印为S 的最小可能值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find minimum startValue
// for positive prefix sum at each index
int minStartValue(vector& nums)
{
    // Store the minimum prefix sum
    int minValue = 0;
 
    // Stores prefix sum at each index
    int sum = 0;
 
    // Traverse over the array
    for (auto n : nums) {
 
        // Update the prefix sum
        sum += n;
 
        // Update the minValue
        minValue = min(minValue, sum);
    }
 
    int startValue = 1 - minValue;
 
    // Return the positive start value
    return startValue;
}
 
// Driver Code
int main()
{
    vector nums = { -3, 2, -3, 4, 2 };
 
    // Function Call
    cout << minStartValue(nums);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
 
class GFG{
 
// Function to find minimum startValue
// for positive prefix sum at each index
static int minStartValue(int[] nums)
{
     
    // Store the minimum prefix sum
    int minValue = 0;
  
    // Stores prefix sum at each index
    int sum = 0;
     
    // Traverse over the array
    for(int n : nums)
    {
         
        // Update the prefix sum
        sum += n;
         
        // Update the minValue
        minValue = Math.min(minValue, sum);
    }
     
    int startValue = 1 - minValue;
     
    // Return the positive start value
    return startValue;
}
  
// Driver Code
public static void main(String[] args)
{
    int[] nums = { -3, 2, -3, 4, 2 };
  
    // Function Call
    System.out.println(minStartValue(nums));
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the
# above approach
 
# Function to find minimum startValue
# for positive prefix sum at each index
def minStartValue(nums):
   
    # Store the minimum prefix sum
    minValue = 0
 
    # Stores prefix sum at each index
    sum = 0
 
    # Traverse over the array
    for i in range(len(nums)):
       
        # Update the prefix sum
        sum += nums[i]
 
        # Update the minValue
        minValue = min(minValue, sum)
 
    startValue = 1 - minValue
     
    # Return the positive start value
    return startValue
 
# Driver Code
if __name__ == '__main__':
   
    nums = [ -3, 2, -3, 4, 2 ]
     
    # Function Call
    print(minStartValue(nums))
 
# This code is contributed by Princi Singh

C#

// C# program for the above approach
using System;
  
class GFG{
   
// Function to find minimum startValue
// for positive prefix sum at each index
static int minStartValue(int[] nums)
{
      
    // Store the minimum prefix sum
    int minValue = 0;
   
    // Stores prefix sum at each index
    int sum = 0;
      
    // Traverse over the array
    foreach(int n in nums)
    {
          
        // Update the prefix sum
        sum += n;
          
        // Update the minValue
        minValue = Math.Min(minValue, sum);
    }
      
    int startValue = 1 - minValue;
      
    // Return the positive start value
    return startValue;
}
   
// Driver Code
public static void Main()
{
    int[] nums = { -3, 2, -3, 4, 2 };
   
    // Function Call
    Console.WriteLine(minStartValue(nums));
}
}
 
// This code is contributed by code_hunt

Javascript

// Javascript program for the above approach
 
// Function to find minimum startValue
// for positive prefix sum at each index
function minStartValue(nums)
{
 
    // Store the minimum prefix sum
    let minValue = 0;
 
    // Stores prefix sum at each index
    let sum = 0;
 
    // Traverse over the array
    for (let n = 0; n < nums.length; n++) {
 
        // Update the prefix sum
        sum += nums[n];
 
        // Update the minValue
        minValue = Math.min(minValue, sum);
    }
    let startValue = 1 - minValue;
 
    // Return the positive start value
    return startValue;
}
 
// Driver Code
    let nums = [ -3, 2, -3, 4, 2 ];
 
    // Function Call
    document.write(minStartValue(nums));
     
    // This code is contributed by souravmahato348.

输出：

5

时间复杂度： O(N)
辅助空间： O(1)

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