Queries for decimal values of subarrays of a binary array

Given a binary array arr[], we to find the number represented by the subarray a[l..r]. There are multiple such queries.

Examples:

Input :  arr[] = {1, 0, 1, 0, 1, 1};
         l = 2, r = 4
         l = 4, r = 5
Output : 5
         3 
Subarray 2 to 4 is 101 which is 5 in decimal.
Subarray 4 to 5 is 11 which is 3 in decimal.

Input : arr[] = {1, 1, 1}
        l = 0, r = 2
        l = 1, r = 2
Output : 7
         3

A Simple Solution is to compute decimal value for every given range using simple binary to decimal conversion. Here each query takes O(len) time where len is length of range.

An Efficient Solution is to do per-computations, so that queries can be answered in O(1) time. 

The number represented by subarray arr[l..r] is arr[l]*+ arr[l+1]*….. + arr[r]*

1. Make an array pre[] of same size as of given array where pre[i] stores the sum of arr[j]*where j includes each value from i to n-1.
2. The number represented by subarray arr[l..r] will be equal to (pre[l] – pre[r+1])/.pre[l] – pre[r+1] is equal to arr[l]*+ arr[l+1]*+……arr[r]*. So if we divide it by , we get the required answer

Flowchart

Flowchart

Implementation:

C++

// C++ implementation of finding number
// represented by binary subarray
#include <bits/stdc++.h>
using namespace std;

// Fills pre[]
void precompute(int arr[], int n, int pre[])
{
    memset(pre, 0, n * sizeof(int));
    pre[n - 1] = arr[n - 1] * pow(2, 0);
    for (int i = n - 2; i >= 0; i--)
        pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));
}

// returns the number represented by a binary
// subarray l to r
int decimalOfSubarr(int arr[], int l, int r,
                    int n, int pre[])
{
    // if r is equal to n-1 r+1 does not exist
    if (r != n - 1)
        return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));

    return pre[l] / (1 << (n - 1 - r));
}

// Driver Function
int main()
{
    int arr[] = { 1, 0, 1, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int pre[n];
    precompute(arr, n, pre);
    cout << decimalOfSubarr(arr, 2, 4, n, pre) << endl;
    cout << decimalOfSubarr(arr, 4, 5, n, pre) << endl;
    return 0;
}

Java

// Java implementation of finding number
// represented by binary subarray
import java.util.Arrays;

class GFG {

    // Fills pre[]
    static void precompute(int arr[], int n, int pre[])
    {
        Arrays.fill(pre, 0);

        pre[n - 1] = arr[n - 1] * (int)(Math.pow(2, 0));
        for (int i = n - 2; i >= 0; i--)
            pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));
    }

    // returns the number represented by a binary
    // subarray l to r
    static int decimalOfSubarr(int arr[], int l, int r,
                               int n, int pre[])
    {

        // if r is equal to n-1 r+1 does not exist
        if (r != n - 1)
            return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));

        return pre[l] / (1 << (n - 1 - r));
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 0, 1, 0, 1, 1 };
        int n = arr.length;

        int pre[] = new int[n];
        precompute(arr, n, pre);

        System.out.println(decimalOfSubarr(arr,
                                           2, 4, n, pre));

        System.out.println(decimalOfSubarr(arr,
                                           4, 5, n, pre));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# implementation of finding number
# represented by binary subarray
from math import pow

# Fills pre[]
def precompute(arr, n, pre):

    pre[n - 1] = arr[n - 1] * pow(2, 0)
    i = n - 2
    while(i >= 0):
        pre[i] = (pre[i + 1] + arr[i] *
                 (1 << (n - 1 - i)))
        i -= 1

# returns the number represented by
# a binary subarray l to r
def decimalOfSubarr(arr, l, r, n, pre):

    # if r is equal to n-1 r+1 does not exist
    if (r != n - 1):
        return ((pre[l] - pre[r + 1]) /
                (1 << (n - 1 - r)))

    return pre[l] / (1 << (n - 1 - r))

# Driver Code
if __name__ == '__main__':
    arr = [1, 0, 1, 0, 1, 1]
    n = len(arr)

    pre = [0 for i in range(n)]
    precompute(arr, n, pre)
    print(int(decimalOfSubarr(arr, 2, 4, n, pre)))
    print(int(decimalOfSubarr(arr, 4, 5, n, pre)))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of finding number
// represented by binary subarray
using System;

class GFG {

    // Fills pre[]
    static void precompute(int[] arr, int n, int[] pre)
    {
        for (int i = 0; i < n; i++)
            pre[i] = 0;

        pre[n - 1] = arr[n - 1] * (int)(Math.Pow(2, 0));
        for (int i = n - 2; i >= 0; i--)
            pre[i] = pre[i + 1] + arr[i] * (1 << (n - 1 - i));
    }

    // returns the number represented by
    // a binary subarray l to r
    static int decimalOfSubarr(int[] arr, int l, int r,
                                      int n, int[] pre)
    {
        // if r is equal to n-1 r+1 does not exist
        if (r != n - 1)
            return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));

        return pre[l] / (1 << (n - 1 - r));
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 0, 1, 0, 1, 1 };
        int n = arr.Length;

        int[] pre = new int[n];
        precompute(arr, n, pre);

        Console.WriteLine(decimalOfSubarr(arr,
                                        2, 4, n, pre));

        Console.WriteLine(decimalOfSubarr(arr,
                                        4, 5, n, pre));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP implementation of finding number
// represented by binary subarray

// Fills pre[]
function precompute(&$arr, $n, &$pre)
{
    $pre[$n - 1] = $arr[$n - 1] * pow(2, 0);
    for ($i = $n - 2; $i >= 0; $i--)
        $pre[$i] = $pre[$i + 1] + $arr[$i] *
                           (1 << ($n - 1 - $i));
}

// returns the number represented by
// a binary subarray l to r
function decimalOfSubarr(&$arr, $l, $r, $n, &$pre)
{
    // if r is equal to n-1 r+1 does not exist
    if ($r != $n - 1)
        return ($pre[$l] - $pre[$r + 1]) /
                    (1 << ($n - 1 - $r));

    return $pre[$l] / (1 << ($n - 1 - $r));
}

// Driver Code
$arr = array(1, 0, 1, 0, 1, 1 );
$n = sizeof($arr);

$pre = array_fill(0, $n, NULL);
precompute($arr, $n, $pre);
echo decimalOfSubarr($arr, 2, 4, $n, $pre) . "\n";
echo decimalOfSubarr($arr, 4, 5, $n, $pre) . "\n";

// This code is contributed by ita_c
?>

Javascript

<script>

// Javascript implementation of finding number
// represented by binary subarray


// Fills pre[]
function precompute(arr, n, pre)
{
    for (let i = 0; i < n; i++)
        pre[i] = 0;

    pre[n - 1] = arr[n - 1] * (Math.pow(2, 0));
    for (let i = n - 2; i >= 0; i--)
        pre[i] = pre[i + 1] + arr[i] *
                 (1 << (n - 1 - i));
}

// returns the number represented by
// a binary subarray l to r
function decimalOfSubarr(arr, l, r,n, pre)
{
    // if r is equal to n-1 r+1 does not exist
    if (r != n - 1)
        return (pre[l] - pre[r + 1]) / (1 << (n - 1 - r));

    return pre[l] / (1 << (n - 1 - r));
}

// Driver code

let arr = [1, 0, 1, 0, 1, 1];
let n = arr.length;

let pre = new Array(n)
precompute(arr, n, pre);

document.write(decimalOfSubarr(arr,2, 4, n, pre)+"<br>");

document.write(decimalOfSubarr(arr, 4, 5, n, pre));

</script>

Output

5
3

Time complexity: O(n)
Auxiliary Space: O(n)

Efficient approach :

traverse the array from the given start index to end index, multiplying each binary digit with the corresponding power of 2 and adding the result.

Implementation :

C++

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

// function to find the decimal equivalent of the subarray
int decimalOfSubarr(int arr[], int l, int r, int n) {
    int ans = 0, p = 1; // initialize ans and p variables to 0 and 1 respectively
    for (int i = r; i >= l; i--) { // loop through the subarray from r to l
        ans += arr[i] * p; // add the current bit multiplied by its corresponding power of 2 to the ans
        p *= 2; // update the power of 2 for the next bit
    }
    return ans; // return the decimal equivalent of the subarray
}

int main() {
    int arr[] = {1, 0, 1, 0, 1, 1}; // initialize the input array
    int n = sizeof(arr) / sizeof(arr[0]); // calculate the size of the array

    // output the decimal equivalents of the subarrays [2, 4] and [4, 5] of the input array
    cout << decimalOfSubarr(arr, 2, 4, n) << endl; // Output: 5
    cout << decimalOfSubarr(arr, 4, 5, n) << endl; // Output: 3
    return 0; // indicate successful program execution
}
//this code is contributed by bhardwajji

Java

import java.util.*;

public class Main {

    // function to find the decimal equivalent of the subarray
    public static int decimalOfSubarr(int[] arr, int l, int r, int n) {
        int ans = 0, p = 1; // initialize ans and p variables to 0 and 1 respectively
        for (int i = r; i >= l; i--) { // loop through the subarray from r to l
            ans += arr[i] * p; // add the current bit multiplied by its corresponding power of 2 to the ans
            p *= 2; // update the power of 2 for the next bit
        }
        return ans; // return the decimal equivalent of the subarray
    }

    public static void main(String[] args) {
        int[] arr = {1, 0, 1, 0, 1, 1}; // initialize the input array
        int n = arr.length; // calculate the size of the array

        // output the decimal equivalents of the subarrays [2, 4] and [4, 5] of the input array
        System.out.println(decimalOfSubarr(arr, 2, 4, n)); // Output: 5
        System.out.println(decimalOfSubarr(arr, 4, 5, n)); // Output: 3
    }
}

Python3

# Python equivalent of the above Java code

def decimalOfSubarr(arr, l, r, n):
    ans = 0
    p = 1
    # loop through the subarray from r to l
    for i in range(r, l-1, -1):
        ans += arr[i] * p # add the current bit multiplied by its corresponding power of 2 to the ans
        p *= 2 # update the power of 2 for the next bit
    return ans # return the decimal equivalent of the subarray

# driver code
arr = [1, 0, 1, 0, 1, 1] # initialize the input array
n = len(arr) # calculate the size of the array

# output the decimal equivalents of the subarrays [2, 4] and [4, 5] of the input array
print(decimalOfSubarr(arr, 2, 4, n)) # Output: 5
print(decimalOfSubarr(arr, 4, 5, n)) # Output: 3

Javascript

function decimalOfSubarr(arr, l, r, n) {
let ans = 0
let p = 1
// loop through the subarray from r to l
for (let i = r; i >= l; i--) {
ans += arr[i] * p // add the current bit multiplied by its corresponding power of 2 to the ans
p *= 2 // update the power of 2 for the next bit
}
return ans // return the decimal equivalent of the subarray
}

// driver code
let arr = [1, 0, 1, 0, 1, 1] // initialize the input array
let n = arr.length // calculate the size of the array

// output the decimal equivalents of the subarrays [2, 4] and [4, 5] of the input array
console.log(decimalOfSubarr(arr, 2, 4, n)) // Output: 5
console.log(decimalOfSubarr(arr, 4, 5, n)) // Output: 3

C#

// C# implementation of finding number
// represented by binary subarray
using System;

class GFG {

    // returns the number represented by
    // a binary subarray l to r
    static int decimalOfSubarr(int[] arr, int l, int r,
                                      int n)
    {
        int ans = 0, p = 1; // initialize ans and p variables to 0 and 1 respectively
        for (int i = r; i >= l; i--) { // loop through the subarray from r to l
            ans += arr[i] * p; // add the current bit multiplied by its corresponding power of 2 to the ans
            p *= 2; // update the power of 2 for the next bit
        }
        return ans; // return the decimal equivalent of the subarray
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 0, 1, 0, 1, 1 };

        int n = arr.Length;

        Console.WriteLine(decimalOfSubarr(arr,
                                        2, 4, n));

        Console.WriteLine(decimalOfSubarr(arr,
                                        4, 5, n));
    }
}

// This code is contributed by shubhamrajput6156

Output

5
3

Time complexity: O(r-l+1), which is equivalent to the length of the subarray
Auxiliary Space: O(1), as we are not using any additional data structures.

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