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Cost required to make all array elements equal to 1
• Last Updated : 22 Apr, 2021

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:

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

Time Complexity: O(N2)
Auxiliary Space: O(1)

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 ` `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);` `    ``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

 ``

Output:

`9`

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

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