# Bitwise OR of N binary strings

Given an array arr[] of binary strings, the task is to calculate the bitwise OR of all of these strings and print the resultant string.

Examples:

Input: arr[] = {“100”, “1001”, “0011”}
Output 1111
0100 OR 1001 OR 0011 = 1111

Input: arr[] = {“10”, “11”, “1000001”}
Output: 1000011

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

Approach: We can do this by first finding the maximum sized string. We need this since we have to add 0s at the front of the strings whose lengths are less than max size. Then apply OR operation on each bit.
For example, if strings are “100”, “001” and “1111”. Here max size is 4, so we have to add 1 zero on first and second string to make their length 4 and then the OR operation can be performed on each of the bits of the numbers resulting in “0100” OR “0001” OR “1111” = “1111”.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the bitwise OR of ` `// all the binary strings ` `string strBitwiseOR(string* arr, ``int` `n) ` `{ ` `    ``string res; ` ` `  `    ``int` `max_size = INT_MIN; ` ` `  `    ``// Get max size and reverse each string ` `    ``// Since we have to perform OR operation ` `    ``// on bits from right to left ` `    ``// Reversing the string will make it easier ` `    ``// to perform operation from left to right ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``max_size = max(max_size, (``int``)arr[i].size()); ` `        ``reverse(arr[i].begin(), arr[i].end()); ` `    ``} ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Add 0s to the end of strings ` `        ``// if needed ` `        ``string s; ` `        ``for` `(``int` `j = 0; j < max_size - arr[i].size(); j++) ` `            ``s += ``'0'``; ` ` `  `        ``arr[i] = arr[i] + s; ` `    ``} ` ` `  `    ``// Perform OR operation on each bit ` `    ``for` `(``int` `i = 0; i < max_size; i++) { ` `        ``int` `curr_bit = 0; ` `        ``for` `(``int` `j = 0; j < n; j++) ` `            ``curr_bit = curr_bit | (arr[j][i] - ``'0'``); ` ` `  `        ``res += (curr_bit + ``'0'``); ` `    ``} ` ` `  `    ``// Reverse the resultant string ` `    ``// to get the final string ` `    ``reverse(res.begin(), res.end()); ` ` `  `    ``// Return the final string ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string arr[] = { ``"10"``, ``"11"``, ``"1000001"` `}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << strBitwiseOR(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the bitwise OR of ` `// all the binary strings ` `static` `String strBitwiseOR(String[] arr, ``int` `n) ` `{ ` `    ``String res=``""``; ` ` `  `    ``int` `max_size = Integer.MIN_VALUE; ` ` `  `    ``// Get max size and reverse each string ` `    ``// Since we have to perform OR operation ` `    ``// on bits from right to left ` `    ``// Reversing the string will make it easier ` `    ``// to perform operation from left to right ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``max_size = Math.max(max_size, (``int``)arr[i].length()); ` `        ``arr[i] = reverse(arr[i]); ` `    ``} ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// Add 0s to the end of strings ` `        ``// if needed ` `        ``String s=``""``; ` `        ``for` `(``int` `j = ``0``; j < max_size - arr[i].length(); j++) ` `            ``s += ``'0'``; ` ` `  `        ``arr[i] = arr[i] + s; ` `    ``} ` `     `  `    ``// Perform OR operation on each bit ` `    ``for` `(``int` `i = ``0``; i < max_size; i++) ` `    ``{ ` `        ``int` `curr_bit = ``0``; ` `        ``for` `(``int` `j = ``0``; j < n; j++) ` `            ``curr_bit = curr_bit | (arr[j].charAt(i) - ``'0'``); ` ` `  `        ``res += (``char``)(curr_bit + ``'0'``); ` `    ``} ` ` `  `    ``// Reverse the resultant string ` `    ``// to get the final string ` `    ``res = reverse(res); ` ` `  `    ``// Return the final string ` `    ``return` `res; ` `} ` ` `  `static` `String reverse(String input)  ` `{ ` `    ``char``[] temparray = input.toCharArray(); ` `    ``int` `left, right = ``0``; ` `    ``right = temparray.length - ``1``; ` ` `  `    ``for` `(left = ``0``; left < right; left++, right--) ` `    ``{ ` `        ``// Swap values of left and right  ` `        ``char` `temp = temparray[left]; ` `        ``temparray[left] = temparray[right]; ` `        ``temparray[right] = temp; ` `    ``} ` `    ``return` `String.valueOf(temparray); ` `}  ` `     `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String arr[] = { ``"10"``, ``"11"``, ``"1000001"` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(strBitwiseOR(arr, n)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the bitwise OR of ` `# all the binary strings ` `def` `strBitwiseOR(arr, n): ` `    ``res``=``"" ` `    ``max_size ``=` `-``(``2``*``*``32``) ` `     `  `    ``# Get max size and reverse each string ` `    ``# Since we have to perform OR operation ` `    ``# on bits from right to left ` `    ``# Reversing the string will make it easier ` `    ``# to perform operation from left to right ` `    ``for` `i ``in` `range``(n): ` `        ``max_size ``=` `max``(max_size, ``len``(arr[i])) ` `        ``arr[i] ``=` `arr[i][::``-``1``] ` `     `  `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Add 0s to the end of strings ` `        ``# if needed ` `        ``s ``=` `"" ` `        ``for` `j ``in` `range``(max_size ``-` `len``(arr[i])): ` `            ``s ``+``=` `'0'` `         `  `        ``arr[i] ``=` `arr[i] ``+` `s ` `         `  `    ``# Perform OR operation on each bit ` `    ``for` `i ``in` `range``(max_size): ` `        ``curr_bit ``=` `0` `        ``for` `j ``in` `range``(n): ` `            ``curr_bit ``=` `curr_bit | ``ord``(arr[j][i]) ` `         `  `        ``res ``+``=` `chr``(curr_bit) ` `     `  `    ``# Reverse the resultant string ` `    ``# to get the final string ` `    ``res``=``res[::``-``1``] ` ` `  `    ``# Return the final string ` `    ``return` `res ` ` `  ` `  `# Driver code ` `arr ``=` `[``"10"``, ``"11"``, ``"1000001"``] ` `n ``=` `len``(arr) ` `print``(strBitwiseOR(arr, n)) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `  `  `// Function to return the bitwise OR of ` `// all the binary strings ` `static` `String strBitwiseOR(String[] arr, ``int` `n) ` `{ ` `    ``String res=``""``; ` `  `  `    ``int` `max_size = ``int``.MinValue; ` `  `  `    ``// Get max size and reverse each string ` `    ``// Since we have to perform OR operation ` `    ``// on bits from right to left ` `    ``// Reversing the string will make it easier ` `    ``// to perform operation from left to right ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``max_size = Math.Max(max_size, (``int``)arr[i].Length); ` `        ``arr[i] = reverse(arr[i]); ` `    ``} ` `  `  `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `  `  `        ``// Add 0s to the end of strings ` `        ``// if needed ` `        ``String s=``""``; ` `        ``for` `(``int` `j = 0; j < max_size - arr[i].Length; j++) ` `            ``s += ``'0'``; ` `  `  `        ``arr[i] = arr[i] + s; ` `    ``} ` `      `  `    ``// Perform OR operation on each bit ` `    ``for` `(``int` `i = 0; i < max_size; i++) ` `    ``{ ` `        ``int` `curr_bit = 0; ` `        ``for` `(``int` `j = 0; j < n; j++) ` `            ``curr_bit = curr_bit | (arr[j][i] - ``'0'``); ` `  `  `        ``res += (``char``)(curr_bit + ``'0'``); ` `    ``} ` `  `  `    ``// Reverse the resultant string ` `    ``// to get the final string ` `    ``res = reverse(res); ` `  `  `    ``// Return the final string ` `    ``return` `res; ` `} ` `  `  `static` `String reverse(String input)  ` `{ ` `    ``char``[] temparray = input.ToCharArray(); ` `    ``int` `left, right = 0; ` `    ``right = temparray.Length - 1; ` `  `  `    ``for` `(left = 0; left < right; left++, right--) ` `    ``{ ` `        ``// Swap values of left and right  ` `        ``char` `temp = temparray[left]; ` `        ``temparray[left] = temparray[right]; ` `        ``temparray[right] = temp; ` `    ``} ` `    ``return` `String.Join(``""``,temparray); ` `}  ` `      `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String []arr = { ``"10"``, ``"11"``, ``"1000001"` `}; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(strBitwiseOR(arr, n)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

```1000011
```

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