# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

## C++

 `// C++ implementation of finding number ` `// represented by binary subarray ` `#include ` `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); ` ` `  `    ``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

 `= 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 ` `?> `

Output:

```5
3
```

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.