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]*2^{r-l}                + arr[l+1]*2^{r - l - 1}                ..... + arr[r]*2^{r-r}        

1. Make an array pre[] of same size as of given array where pre[i] stores the sum of arr[j]*2^{n - 1 - 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])/2^{n-1-r}                .pre[l] - pre[r+1] is equal to arr[l]*2^{n - 1 - l}                + arr[l+1]*2^{n - 1 - l - 1}                +......arr[r]*2^{n - 1 - r}                . So if we divide it by 2^{n - 1 - r}                , we get the required answer

 Flowchart

Implementation:

// 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 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. 

# 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# 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 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 ?> 

<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> 

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 :

#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 

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  } } 

# 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 

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# 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 

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.