Decimal equivalent of concatenation of absolute difference of floor and rounded-off values of array elements as a Binary String

Given an array arr[] consisting of N floating-point numbers, the task is to print the decimal representation of the binary array constructed from the absolute difference between the floor and round-off value for each array element. 

Examples:

Input: arr[] = {1.2, 2.6, 4.2, 6.9, 3.1, 21.6, 91.2}
Output: 42
Explanation:
Below is the image to illustrate the above example:

Input: arr[] = {5.7, 2.8, 1.9, 5.6, 2.2}
Output: 30

Approach: Follow the steps below to solve the problem:

• Initialize a variable, say result as 0 that stores the resultant numbers formed.
• Initialize a variable, say power as 0 that keeps the power of 2 added in each step.
• Traverse the given array arr[] from the end and perform the following steps:
  • Initialize a variable, say bit that stores the absolute difference between the round-off value and the floor value of each array element.
  • If the value of the absolute difference is 1, then multiply the digit with the proper power of 2 and add it to the variable result.
  • Increment the value of power by 1.
• After completing the above steps, print the value of the result as the required decimal equivalent of the binary representation.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the decimal equivalent
// of the new binary array constructed
// from absolute decimal of floor and
// the round-off values
int findDecimal(float arr[], int N)
{
    int bit, power = 0, result = 0;

    // Traverse the givenarray from
    // the end
    for (int i = N - 1; i >= 0; i--) {

        // Stores the absolute difference
        // between floor and round-off
        // each array element
        bit = abs(floor(arr[i])
                  - round(arr[i]));

        // If bit / difference is 1, then
        // calculate the bit by proper
        // power of 2 and add it to result
        if (bit)
            result += pow(2, power);

        // Increment the value of power
        power++;
    }

    // Print the result
    cout << result;
}

// Driver Code
int main()
{
    float arr[] = { 1.2, 2.6, 4.2, 6.9,
                    3.1, 21.6, 91.2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    findDecimal(arr, N);

    return 0;
}

Java

// Java program for above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG
{

// Function to find the decimal equivalent
// of the new binary array constructed
// from absolute decimal of floor and
// the round-off values
static void findDecimal(double arr[], int N)
{
    int bit, power = 0, result = 0;

    // Traverse the givenarray from
    // the end
    for (int i = N - 1; i >= 0; i--)
    {

        // Stores the absolute difference
        // between floor and round-off
        // each array element
        bit = Math.abs((int)Math.floor(arr[i])
                  - (int)Math.round(arr[i]));

        // If bit / difference is 1, then
        // calculate the bit by proper
        // power of 2 and add it to result
        if (bit != 0)
            result += Math.pow(2, power);

        // Increment the value of power
        power++;
    }

    // Print the result
    System.out.print(result);
}

    // Driver Code
    public static void main(String[] args)
    {
    double arr[] = { 1.2, 2.6, 4.2, 6.9,
                    3.1, 21.6, 91.2 };
    int N = arr.length;
    findDecimal(arr, N);
    }
}

// This code is contributed by souravghosh0416.

Python3

# Python program for the above approach


# Function to find the decimal equivalent
# of the new binary array constructed
# from absolute decimal of floor and
# the round-off values
def findDecimal(arr, N):
    power = 0;
    result = 0;

    # Traverse the givenarray from
    # the end
    for i in range(N - 1, -1, -1):

        # Stores the absolute difference
        # between floor and round-off
        # each array element
        bit = abs(int(arr[i]) - round(arr[i]));

        # If bit / difference is 1, then
        # calculate the bit by proper
        # power of 2 and add it to result
        if (bit):
            result += pow(2, power);

        # Increment the value of power
        power += 1;

    # Print the result
    print(result);


# Driver Code

arr = [ 1.2, 2.6, 4.2, 6.9, 3.1, 21.6, 91.2 ];
N = len(arr)
findDecimal(arr, N);

# This code is contributed by gfgking.

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the decimal equivalent
// of the new binary array constructed
// from absolute decimal of floor and
// the round-off values
static void findDecimal(double[] arr, int N)
{
    int bit, power = 0, result = 0;

    // Traverse the givenarray from
    // the end
    for(int i = N - 1; i >= 0; i--)
    {

        // Stores the absolute difference
        // between floor and round-off
        // each array element
        bit = Math.Abs((int)Math.Floor(arr[i]) -
                       (int)Math.Round(arr[i]));

        // If bit / difference is 1, then
        // calculate the bit by proper
        // power of 2 and add it to result
        if (bit != 0)
            result += (int)Math.Pow(2, power);

        // Increment the value of power
        power++;
    }

    // Print the result
    Console.WriteLine(result);
}

// Driver Code
public static void Main()
{
    double[] arr = { 1.2, 2.6, 4.2, 6.9,
                     3.1, 21.6, 91.2 };
    int N = arr.Length;

    findDecimal(arr, N);
}
}

// This code is contriobuted by sanjoy_62

Javascript

<script>

        // JavaScript program for the above approach

        // Function to find the decimal equivalent
        // of the new binary array constructed
        // from absolute decimal of floor and
        // the round-off values
        function findDecimal(arr, N) {
            let bit, power = 0, result = 0;

            // Traverse the givenarray from
            // the end
            for (let i = N - 1; i >= 0; i--) {

                // Stores the absolute difference
                // between floor and round-off
                // each array element
                bit = Math.abs(Math.floor(arr[i])
                    - Math.round(arr[i]));

                // If bit / difference is 1, then
                // calculate the bit by proper
                // power of 2 and add it to result
                if (bit != 0)
                    result += Math.pow(2, power);

                // Increment the value of power
                power++;
            }

            // Print the result
            document.write(result);
        }

        // Driver Code

        let arr = [1.2, 2.6, 4.2, 6.9,
            3.1, 21.6, 91.2];
        let N = arr.length;
        findDecimal(arr, N);

        // This code is contributed by Hritik

</script>

Output:

42

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