执行给定操作后更新数组的总和

给定N个元素的数组arr [] ，任务是更新所有数组元素，以使元素arr [i]更新为arr [i] = arr [i] – X其中X = arr [i + 1] + arr [i + 2] +…+ arr [N – 1] ，最后打印更新后的数组之和。

例子：

Input: arr[] = {40, 25, 12, 10}
Output: 8
The updated array will be {-7, 3, 2, 10}.
-7 + 3 + 2 + 10 = 8

Input: arr[] = {50, 30, 10, 2, 0}
Output: 36

方法：对于i的每个可能值，一个简单的解决方案是更新arr [i] = arr [i] – sum(arr [i + 1…N-1])。

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Utility function to return
// the sum of the array
int sumArr(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    return sum;
}
  
// Function to return the sum
// of the modified array
int sumModArr(int arr[], int n)
{
  
    for (int i = 0; i < n - 1; i++) {
  
        // Find the sum of the subarray
        // arr[i+1...n-1]
        int subSum = 0;
        for (int j = i + 1; j < n; j++) {
            subSum += arr[j];
        }
  
        // Subtract the subarray sum
        arr[i] -= subSum;
    }
  
    // Return the sum of
    // the modified array
    return sumArr(arr, n);
}
  
// Driver code
int main()
{
    int arr[] = { 40, 25, 12, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << sumModArr(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Utility function to return
// the sum of the array
static int sumArr(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    return sum;
}
  
// Function to return the sum
// of the modified array
static int sumModArr(int arr[], int n)
{
    for (int i = 0; i < n - 1; i++)
    {
  
        // Find the sum of the subarray
        // arr[i+1...n-1]
        int subSum = 0;
        for (int j = i + 1; j < n; j++)
        {
            subSum += arr[j];
        }
  
        // Subtract the subarray sum
        arr[i] -= subSum;
    }
  
    // Return the sum of
    // the modified array
    return sumArr(arr, n);
}
  
// Driver code
public static void main(String []args)
{
    int arr[] = { 40, 25, 12, 10 };
    int n = arr.length;
  
    System.out.println(sumModArr(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
  
# Utility function to return
# the sum of the array
def sumArr(arr, n):
    sum = 0
    for i in range(n):
        sum += arr[i]
    return sum
  
# Function to return the sum
# of the modified array
def sumModArr(arr, n):
  
    for i in range(n - 1):
  
        # Find the sum of the subarray
        # arr[i+1...n-1]
        subSum = 0
        for j in range(i + 1, n):
            subSum += arr[j]
              
        # Subtract the subarray sum
        arr[i] -= subSum
  
    # Return the sum of
    # the modified array
    return sumArr(arr, n)
  
# Driver code
arr = [40, 25, 12, 10]
n = len(arr)
  
print(sumModArr(arr, n))
  
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
    // Utility function to return
    // the sum of the array
    static int sumArr(int []arr, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        return sum;
    }
      
    // Function to return the sum
    // of the modified array
    static int sumModArr(int []arr, int n)
    {
        for (int i = 0; i < n - 1; i++)
        {
      
            // Find the sum of the subarray
            // arr[i+1...n-1]
            int subSum = 0;
            for (int j = i + 1; j < n; j++)
            {
                subSum += arr[j];
            }
      
            // Subtract the subarray sum
            arr[i] -= subSum;
        }
      
        // Return the sum of
        // the modified array
        return sumArr(arr, n);
    }
      
    // Driver code
    public static void Main()
    {
        int []arr = { 40, 25, 12, 10 };
        int n = arr.Length;
      
        Console.WriteLine(sumModArr(arr, n));
    }
}
  
// This code is contributed by AnkitRai01

输出：

8

时间复杂度： O(N ² )

高效的方法：一种有效的解决方案是从末尾遍历数组，以便直到现在为止子数组的总和，即sum(arr [i + 1…n-1])可用于计算当前子数组arr [ i…n-1]即sum(arr [i…n-1])= arr [i] + sum(arr [i + 1…n-1]) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Utility function to return
// the sum of the array
int sumArr(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    return sum;
}
  
// Function to return the sum
// of the modified array
int sumModArr(int arr[], int n)
{
  
    int subSum = arr[n - 1];
    for (int i = n - 2; i >= 0; i--) {
  
        int curr = arr[i];
  
        // Subtract the subarray sum
        arr[i] -= subSum;
  
        // Sum of subarray arr[i...n-1]
        subSum += curr;
    }
  
    // Return the sum of
    // the modified array
    return sumArr(arr, n);
}
  
// Driver code
int main()
{
    int arr[] = { 40, 25, 12, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << sumModArr(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
      
    // Utility function to return 
    // the sum of the array 
    static int sumArr(int arr[], int n) 
    { 
        int sum = 0; 
        for (int i = 0; i < n; i++) 
            sum += arr[i]; 
        return sum; 
    } 
      
    // Function to return the sum 
    // of the modified array 
    static int sumModArr(int arr[], int n) 
    { 
        int subSum = arr[n - 1]; 
        for (int i = n - 2; i >= 0; i--)
        { 
            int curr = arr[i]; 
      
            // Subtract the subarray sum 
            arr[i] -= subSum; 
      
            // Sum of subarray arr[i...n-1] 
            subSum += curr; 
        } 
      
        // Return the sum of 
        // the modified array 
        return sumArr(arr, n); 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    { 
        int []arr = { 40, 25, 12, 10 }; 
        int n = arr.length; 
      
        System.out.println(sumModArr(arr, n)); 
    } 
}
  
// This code is contributed by kanugargng

Python3

# Python3 implementation of the approach
  
# Utility function to return
# the sum of the array
def sumArr(arr, n):
  
    sum = 0;
    for i in range(n):
        sum += arr[i];
    return sum;
  
# Function to return the sum
# of the modified array
def sumModArr(arr, n):
  
    subSum = arr[n - 1];
    for i in range(n - 2, -1, -1):
  
        curr = arr[i];
  
        # Subtract the subarray sum
        arr[i] -= subSum;
  
        # Sum of subarray arr[i...n-1]
        subSum += curr;
  
    # Return the sum of
    # the modified array
    return sumArr(arr, n);
  
# Driver code
arr = [40, 25, 12, 10 ];
n = len(arr);
  
print(sumModArr(arr, n));
  
# This code is contributed by 29AjayKumar

C＃

// C# implementation of the approach
using System;
      
class GFG 
{
      
    // Utility function to return 
    // the sum of the array 
    static int sumArr(int []arr, int n) 
    { 
        int sum = 0; 
        for (int i = 0; i < n; i++) 
            sum += arr[i]; 
        return sum; 
    } 
      
    // Function to return the sum 
    // of the modified array 
    static int sumModArr(int []arr, int n) 
    { 
        int subSum = arr[n - 1]; 
        for (int i = n - 2; i >= 0; i--)
        { 
            int curr = arr[i]; 
      
            // Subtract the subarray sum 
            arr[i] -= subSum; 
      
            // Sum of subarray arr[i...n-1] 
            subSum += curr; 
        } 
      
        // Return the sum of 
        // the modified array 
        return sumArr(arr, n); 
    } 
      
    // Driver code 
    public static void Main (String[] args) 
    { 
        int []arr = { 40, 25, 12, 10 }; 
        int n = arr.Length; 
      
        Console.WriteLine(sumModArr(arr, n)); 
    } 
}
  
// This code is contributed by 29AjayKumar

输出：

8

时间复杂度： O(N)