Given two binary numbers strings 
and 
of length 
. Find the number of ways of swapping two bits in s1(only s1 not s2) so that bit-wise OR of these two numbers s1 and s2 are changed.
Note: The length of both string must be equal, you can take leading zeros in case of different length.
Example:
Input: s1 = "01011", s2 = "11001" Output: 4 Explanation: You can swap the bit of s1 at indexed: (1, 4), (2, 3), (3, 4) and (3, 5) there are 4 ways possible. Input: s1 = "011000", s2 = "010011" Output: 6
Approach: Initialize a variable result as zero that stores number of ways to swap bits of s1 so that bitwise OR of s1 and s2 changes. Initialize four variables say 
, 
, 
, and 
as zero.
Traverse both strings from left side and check for the below conditions to increment values of the
variable declared above:
- If the current bit of both s1 and s2 are zero, increment c.
- If current bit of s2 is zero and s1 is one, increment d.
- If current bit of s2 is one and s1 is zero, increment a.
- If current bit of both s1 and s2 is one, increment b.
Update the result by (a*d) + (b*c) + (c*d) and return the result.
Below is the implementation of the above approach:
C++
// C++ program to find no of ways// to swap bits of s1 so that// bitwise OR of s1 and s2 changes#include <iostream>using namespace std;// Function to find number of waysint countWays(string s1, string s2, int n){ int a, b, c, d; a = b = c = d = 0; // initialise result that store // No. of swaps required int result = 0; // Traverse both strings and check // the bits as explained for (int i = 0; i < n; i++) { if (s2[i] == '0') { if (s1[i] == '0') { c++; } else { d++; } } else { if (s1[i] == '0') { a++; } else { b++; } } } // calculate result result = a * d + b * c + c * d; return result;}// Driver codeint main(){ int n = 5; string s1 = "01011"; string s2 = "11001"; cout << countWays(s1, s2, n); return 0;} |
Java
// Java program to find no of ways// to swap bits of s1 so that// bitwise OR of s1 and s2 changesimport java.io.*;class GFG { // Function to find number of waysstatic int countWays(String s1, String s2, int n){ int a, b, c, d; a = b = c = d = 0; // initialise result that store // No. of swaps required int result = 0; // Traverse both strings and check // the bits as explained for (int i = 0; i < n; i++) { if (s2.charAt(i) == '0') { if (s1.charAt(i) == '0') { c++; } else { d++; } } else { if (s1.charAt(i) == '0') { a++; } else { b++; } } } // calculate result result = a * d + b * c + c * d; return result;}// Driver code public static void main (String[] args) { int n = 5; String s1 = "01011"; String s2 = "11001"; System.out.println(countWays(s1, s2, n)); }}// This code is contributed by shs.. |
Python3
# Python 3 program to find no of ways# to swap bits of s1 so that# bitwise OR of s1 and s2 changes# Function to find number of waysdef countWays(s1, s2, n): a = b = c = d = 0 # initialise result that store # No. of swaps required result = 0; # Traverse both strings and check # the bits as explained for i in range(0, n, 1): if (s2[i] == '0'): if (s1[i] == '0'): c += 1; else: d += 1 else: if (s1[i] == '0'): a += 1 else: b += 1 # calculate result result = a * d + b * c + c * d return result# Driver codeif __name__ == '__main__': n = 5 s1 = "01011" s2 = "11001" print(countWays(s1, s2, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find no of ways// to swap bits of s1 so that// bitwise OR of s1 and s2 changesusing System;class GFG {// Function to find number of waysstatic int countWays(string s1, string s2, int n){ int a, b, c, d; a = b = c = d = 0; // initialise result that store // No. of swaps required int result = 0; // Traverse both strings and check // the bits as explained for (int i = 0; i < n; i++) { if (s2[i] == '0') { if (s1[i] == '0') { c++; } else { d++; } } else { if (s1[i] == '0') { a++; } else { b++; } } } // calculate result result = a * d + b * c + c * d; return result;}// Driver code public static void Main () { int n = 5; string s1 = "01011"; string s2 = "11001"; Console.WriteLine(countWays(s1, s2, n)); }}// This code is contributed by shs.. |
PHP
<?php// PHP program to find no of ways// to swap bits of s1 so that// bitwise OR of s1 and s2 changes// Function to find number of waysfunction countWays($s1, $s2, $n){ $a = $b = $c = $d = 0; // initialise result that store // No. of swaps required $result = 0; // Traverse both strings and check // the bits as explained for ($i = 0; $i < $n; $i++) { if ($s2[$i] == '0') { if ($s1[$i] == '0') { $c++; } else { $d++; } } else { if ($s1[$i] == '0') { $a++; } else { $b++; } } } // calculate result $result = $a * $d + $b * $c + $c * $d; return $result;}// Driver code$n = 5;$s1 = "01011";$s2 = "11001";echo countWays($s1, $s2, $n);// This code is contributed by ajit?> |
Javascript
<script>// javascript program to find no of ways// to swap bits of s1 so that// bitwise OR of s1 and s2 changes // Function to find number of waysfunction countWays(s1, s2 , n){ var a, b, c, d; a = b = c = d = 0; // initialise result that store // No. of swaps required var result = 0; // Traverse both strings and check // the bits as explained for (i = 0; i < n; i++) { if (s2.charAt(i) == '0') { if (s1.charAt(i) == '0') { c++; } else { d++; } } else { if (s1.charAt(i) == '0') { a++; } else { b++; } } } // calculate result result = a * d + b * c + c * d; return result;}// Driver codevar n = 5;var s1 = "01011";var s2 = "11001";document.write(countWays(s1, s2, n));// This code is contributed by Princi Singh </script> |
Output
4
Complexity y Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
