使所有数组元素等于1所需的成本

给定大小为N的二进制数组arr [] ，任务是查找使所有数组元素等于1所需的总成本，其中将任何0转换为1的成本等于在该0之前存在的1的计数。

例子：

Input: arr[] = {1, 0, 1, 0, 1, 0}
Output: 9
Explanation:
Following operations are performed:

1. Converting arr[1] to 1 modifies arr[] to {1, 1, 1, 0, 1, 0}. Cost = 1.
2. Converting arr[3] to 1 modifies arr[] to {1, 1, 1, 1, 1, 0}. Cost = 3.
3. Converting arr[5] to 1 modifies arr[] to {1, 1, 1, 1, 1, 5}. Cost = 5.

Therefore, the total cost is 1 + 3 + 5 = 9.

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

天真的方法：解决给定问题的最简单方法是遍历数组arr []并计算每个包含0的索引之前存在的1 s的数量，并打印所有获得的成本之和。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：可以通过观察以下事实来优化上述方法：在将每个0转换为1之后，在每个0之前存在的1 s的计数由发生0的索引给出。因此，任务是遍历给定数组并在数组arr []中打印所有具有0 s的索引的所有和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to calculate the cost required
// to make all array elements equal to 1
int findCost(int A[], int N)
{
    // Stores the total cost
    int totalCost = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // If current element is 0
        if (A[i] == 0) {
 
            // Convert 0 to 1
            A[i] = 1;
 
            // Add the cost
            totalCost += i;
        }
    }
 
    // Return the total cost
    return totalCost;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 0, 1, 0, 1, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << findCost(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
 
class GFG{
 
// Function to calculate the cost required
// to make all array elements equal to 1
static int findCost(int[] A, int N)
{
     
    // Stores the total cost
    int totalCost = 0;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // If current element is 0
        if (A[i] == 0)
        {
             
            // Convert 0 to 1
            A[i] = 1;
 
            // Add the cost
            totalCost += i;
        }
    }
 
    // Return the total cost
    return totalCost;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 0, 1, 0, 1, 0 };
    int N = arr.length;
 
    System.out.println(findCost(arr, N));
}
}
 
// This code is contributed by ukasp

Python3

# Python3 program for the above approach
 
# Function to calculate the cost required
# to make all array elements equal to 1
def findCost(A, N):
 
    # Stores the total cost
    totalCost = 0
 
    # Traverse the array arr[]
    for i in range(N):
 
        # If current element is 0
        if (A[i] == 0):
 
            # Convert 0 to 1
            A[i] = 1
 
            # Add the cost
            totalCost += i
 
    # Return the total cost
    return totalCost
 
# Driver Code
if __name__ == '__main__':
 
    arr = [ 1, 0, 1, 0, 1, 0 ]
    N = len(arr)
     
    print(findCost(arr, N))
     
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to calculate the cost required
// to make all array elements equal to 1
static int findCost(int []A, int N)
{
     
    // Stores the total cost
    int totalCost = 0;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // If current element is 0
        if (A[i] == 0)
        {
 
            // Convert 0 to 1
            A[i] = 1;
 
            // Add the cost
            totalCost += i;
        }
    }
 
    // Return the total cost
    return totalCost;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 1, 0, 1, 0, 1, 0 };
    int N = arr.Length;
     
    Console.Write(findCost(arr, N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

 
 
// Javascript program for the above approach
 
// Function to calculate the cost required
// to make all array elements equal to 1
function findCost(A, N)
{
    // Stores the total cost
    var totalCost = 0;
 
    var i;
    // Traverse the array arr[]
    for (i = 0; i < N; i++) {
 
        // If current element is 0
        if (A[i] == 0) {
 
            // Convert 0 to 1
            A[i] = 1;
 
            // Add the cost
            totalCost += i;
        }
    }
 
    // Return the total cost
    return totalCost;
}
 
// Driver Code
    var arr = [1, 0, 1, 0, 1, 0]
    var N = arr.length
    document.write(findCost(arr, N));
 

输出：

9

时间复杂度： O(N)
辅助空间： O(1)