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Bitwise OR of N binary strings in C++
In this problem, we are given an array bin[] of size n of binary strings. Our task is to create a program to find the Bitwise OR (&) of n binary strings.
Here, we will take all numbers and find the bitwise AND of them i.e. bin[0] | bin[1] |... bin[n-2] | bin[n]
Let’s take an example to understand the problem,
Input −
bin[] = {“1001”, “11001”, “010101”}
Output −
011101
Explanation − Bitwise OR of all binary string −
(1001) | (11001) | (010101) = 011101
To solve this problem, We will simply find the string with the largest number of bits (max length string). Then we will add an adequate number of leading 0 to all strings. Then find the Bitwise OR of the bits.
Let’s take an example to show the working of the algorithm −
bin[] = {“1101”, “011010” , “00111”}
Max length string is 011010 with length 6. So, we will add leading 0’s to other strings.
Updated strings − “001101”, “011010”, “000111”.
Finding BITWISE OR of all strings − 001101 | 011010 | 000111 = 011111
Example
Program to illustrate the working of our solution −
#include <bits/stdc++.h> using namespace std; string bitwiseOR(string* bin, int n){ string result; int max_size = INT_MIN; for (int i = 0; i < n; i++) { max_size = max(max_size, (int)bin[i].size()); reverse(bin[i].begin(), bin[i].end()); } for (int i = 0; i < n; i++) { string s; for (int j = 0; j < max_size - bin[i].size(); j++) s += '0'; bin[i] = bin[i] + s; } for (int i = 0; i < max_size; i++) { int insertBit = 0; for (int j = 0; j < n; j++) insertBit = insertBit | (bin[j][i] - '0'); result += (insertBit + '0'); } reverse(result.begin(), result.end()); return result; } int main() { string bin[] = { "1101", "011010", "00111" }; int n = sizeof(bin) / sizeof(bin[0]); cout<<"The bitwise OR of all the binary String of the string array is "<<bitwiseOR(bin, n); return 0; }
Output
The bitwise OR of all the binary String of the string array is 011111