最小化使数组元素的和与乘积不为零所需的增量或减量

给定N个整数的数组arr [] ，任务是计算数组上所需的最小递增或递减运算数，以使数组arr []的所有元素的和与乘积不为零。

例子：

Input: arr[] = {-1, -1, 0, 0}
Output: 2
Explanation: Perform the following operations to update the array as:
Operation 1: Incrementing arr[2] modifies arry to {-1, -1, 1, 0}.
Operation 2: Decrementing arr[3] modifies array to {-1, -1, 1, -1}.
Therefore, the sum and product of the above array is -2 and -1 which is non-zero.

Input: arr[] = {-2, 1, 0}
Output: 1

方法：可以根据以下观察结果解决给定问题：

• 使数组乘积为非零以及为乘积为非零所需的最少步骤，所有元素都必须为非零。
• 如果总和为负，则使数组总和为非零所需的最少步骤，然后将所有0s元素减1 ，如果总和为正，则将所有零元素增加1 ，如果总和为非零，则，只需增加或减少数组的任何元素。

请按照以下步骤解决此问题：

1. 遍历给定的数组并计算数组中的零个数。
2. 找到给定数组的总和。
3. 如果零计数大于0，则结果为该计数。
4. 否则，如果总和等于0，则结果为1 。
5. 否则结果将为0 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the sum of array
int array_sum(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    // Return the sum
    return sum;
}
 
// Function that counts the minimum
// operations required to make the
// sum and product of array non-zero
int countOperations(int arr[], int N)
{
    // Stores count of zero elements
    int count_zeros = 0;
 
    // Iterate over the array to
    // count zero elements
    for (int i = 0; i < N; i++) {
        if (arr[i] == 0)
            count_zeros++;
    }
 
    // Sum of elements of the array
    int sum = array_sum(arr, N);
 
    // Print the result
    if (count_zeros)
        return count_zeros;
 
    if (sum == 0)
        return 1;
 
    return 0;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { -1, -1, 0, 0 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << countOperations(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
   
// Function to find the
// sum of array
static int array_sum(int arr[],
                     int n)
{
  int sum = 0;
   
  for (int i = 0; i < n; i++)
    sum += arr[i];
 
  // Return the sum
  return sum;
}
 
// Function that counts the minimum
// operations required to make the
// sum and product of array non-zero
static int countOperations(int arr[],
                           int N)
{
  // Stores count of zero
  // elements
  int count_zeros = 0;
 
  // Iterate over the array to
  // count zero elements
  for (int i = 0; i < N; i++)
  {
    if (arr[i] == 0)
      count_zeros++;
  }
 
  // Sum of elements of the
  // array
  int sum = array_sum(arr, N);
 
  // Print the result
  if (count_zeros != 0)
    return count_zeros;
 
  if (sum == 0)
    return 1;
 
  return 0;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {-1, -1, 0, 0};
 
  // Size of array
  int N = arr.length;
 
  // Function Call
  System.out.print(countOperations(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the
# above approach
 
# Function to find the
# sum of array
def array_sum(arr, n):
   
    sum = 0
     
    for i in range(n):
        sum += arr[i]
         
    # Return the sum
    return sum
 
# Function that counts the minimum
# operations required to make the
# sum and product of array non-zero
def countOperations(arr, N):
   
    # Stores count of zero
    # elements
    count_zeros = 0
 
    # Iterate over the array to
    # count zero elements
    for i in range(N):
        if (arr[i] == 0):
            count_zeros+=1
 
    # Sum of elements of the
    # array
    sum = array_sum(arr, N)
 
    # Prthe result
    if (count_zeros):
        return count_zeros
 
    if (sum == 0):
        return 1
 
    return 0
 
# Driver Code
if __name__ == '__main__':
   
    # Given array arr[]
    arr = [-1, -1, 0, 0]
 
    # Size of array
    N = len(arr)
 
    # Function Call
    print(countOperations(arr, N))
 
# This code is contributed by Mohit Kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the
// sum of array
static int array_sum(int[] arr,
                     int n)
{
  int sum = 0;
   
  for(int i = 0; i < n; i++)
    sum += arr[i];
 
  // Return the sum
  return sum;
}
 
// Function that counts the minimum
// operations required to make the
// sum and product of array non-zero
static int countOperations(int[] arr,
                           int N)
{
   
  // Stores count of zero
  // elements
  int count_zeros = 0;
 
  // Iterate over the array to
  // count zero elements
  for(int i = 0; i < N; i++)
  {
    if (arr[i] == 0)
      count_zeros++;
  }
 
  // Sum of elements of the
  // array
  int sum = array_sum(arr, N);
 
  // Print the result
  if (count_zeros != 0)
    return count_zeros;
 
  if (sum == 0)
    return 1;
 
  return 0;
}
 
// Driver Code
public static void Main()
{
   
  // Given array arr[]
  int[] arr = { -1, -1, 0, 0 };
 
  // Size of array
  int N = arr.Length;
 
  // Function call
  Console.Write(countOperations(arr, N));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// Javascript program for the above approach
 
    // Function to find the
    // sum of array
    function array_sum(arr , n) {
        var sum = 0;
 
        for (i = 0; i < n; i++)
            sum += arr[i];
 
        // Return the sum
        return sum;
    }
 
    // Function that counts the minimum
    // operations required to make the
    // sum and product of array non-zero
    function countOperations(arr , N) {
        // Stores count of zero
        // elements
        var count_zeros = 0;
 
        // Iterate over the array to
        // count zero elements
        for (i = 0; i < N; i++) {
            if (arr[i] == 0)
                count_zeros++;
        }
 
        // Sum of elements of the
        // array
        var sum = array_sum(arr, N);
 
        // Prvar the result
        if (count_zeros != 0)
            return count_zeros;
 
        if (sum == 0)
            return 1;
 
        return 0;
    }
 
    // Driver Code
     
        // Given array arr
        var arr = [ -1, -1, 0, 0 ];
 
        // Size of array
        var N = arr.length;
 
        // Function Call
        document.write(countOperations(arr, N));
 
// This code contributed by umadevi9616
 

输出：

2

时间复杂度： O(N)
辅助空间： O(1)