Cost required to make all array elements equal to 1

Last Updated : 20 Oct, 2023

Given a binary array arr[] of size N, the task is to find the total cost required to make all array elements equal to 1, where the cost of converting any 0 to 1 is equal to the count of 1s present before that 0.

Examples:

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

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

Naive Approach: The simplest approach to solve the given problem is to traverse the array arr[] and count the numbers of 1s present before every index containing 0 and print the sum of all the costs obtained.

Implementation:

C++

#include <iostream>
using namespace std;

int getTotalCost(int arr[],int N) {
   int countones = 0;
  int cost=0;
   for (int i = 0; i <N; i++) {
       if(arr[i]==1)
       {
           countones++;
           continue;
       }
       if (arr[i] == 0) {
           cost += countones;
         countones++;
       }
   }
   return cost;
}

int main() {
   int arr[] = {1,0,1,0,1,0};

   int N = sizeof(arr) / sizeof(arr[0]);
   cout <<getTotalCost(arr,N)<<endl;

   return 0;
}

Java

public class TotalCost {

    public static int getTotalCost(int[] arr) {
        int countOnes = 0;
        int cost = 0;

        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == 1) {
                countOnes++;
            } else if (arr[i] == 0) {
                cost += countOnes;
                countOnes++;
            }
        }

        return cost;
    }

    public static void main(String[] args) {
        int[] arr = {1, 0, 1, 0, 1, 0};

        System.out.println(getTotalCost(arr));
    }
}

Python3

def get_total_cost(arr):
    count_ones = 0
    cost = 0
    for num in arr:
        if num == 1:
            count_ones += 1
            continue
        if num == 0:
            cost += count_ones
            count_ones += 1
    return cost

if __name__ == "__main__":
    arr = [1, 0, 1, 0, 1, 0]
    print(get_total_cost(arr))

#Contributed by Aditi Tyagi

using System;

class Program
{
    // Function to calculate the total cost
    static int GetTotalCost(int[] arr)
    {
        int countOnes = 0; // Initialize a counter for '1's
        int cost = 0; // Initialize the total cost

        for (int i = 0; i < arr.Length; i++)
        {
            if (arr[i] == 1)
            {
                countOnes++; // Increment the count of '1's
                continue; // Skip to the next element
            }
            if (arr[i] == 0)
            {
                cost += countOnes; // Increment the cost by the current count of '1's
                countOnes++; // Increment the count of '1's for the next element
            }
        }

        return cost; // Return the total cost
    }

    static void Main(string[] args)
    {
        int[] arr = { 1, 0, 1, 0, 1, 0 }; // Define the array of 0s and 1s

        // Calculate and print the total cost
        Console.WriteLine("Total Cost: " + GetTotalCost(arr));
    }
}

JavaScript

// Define a function to calculate the total cost
function getTotalCost(arr) {
   let countOnes = 0;
   let cost = 0;

   // Loop through the array
   for (let i = 0; i < arr.length; i++) {
       // If the current element is 1, increment the count of ones
       if (arr[i] === 1) {
           countOnes++;
           continue; // Skip the rest of the loop iteration
       }

       // If the current element is 0, calculate and update the cost
       if (arr[i] === 0) {
           cost += countOnes;
           countOnes++;
       }
   }

   return cost;
}

// Define the main function
function main() {
   const arr = [1, 0, 1, 0, 1, 0];

   // Calculate the length of the array
   const N = arr.length;

   // Call the getTotalCost function and print the result
   console.log(getTotalCost(arr));
}

// Call the main function to start the program
main();

Output

Time Complexity: O(N), where N is the size of the array.
Auxiliary Space: O(1), as we are not using any extra array.

Efficient Approach: The above approach can be optimized by observing the fact that after converting every 0 to 1, the count of 1s present before every 0 is given by the index at which 0 occurs. Therefore, the task is to traverse the given array and print all the sum of all the indices having 0s in the array arr[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
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# 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

<script>


// 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));

</script>

Output

Time Complexity: O(N)
Auxiliary Space: O(1)

Minimum operations required to make all the array elements equal

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