Check if two binary strings can be made equal by swapping 1s occurring before 0s
Given two binary strings str1 and str2 having same length, the task is to find if it is possible to make the two binary strings str1 and str2 equal by swapping all 1s occurring at indices less than that of 0s index in the binary string str1.
Examples:
Input: str1 = “0110”, str2 = “0011”
Output: Possible
Explanation:
Swapping str1[2] with str1[3] the binary string str1 becomes “0101”.
Swapping str1[1] with str1[2] the binary string str1 becomes “0011” .
The binary string str1 becomes equal to the binary string str2 therefore, the required output is Possible.Input: str1 = “101”, str2 = “010”
Output: Not Possible
Approach: The idea is to count the number of 1s and 0s in str1 and str2 and then proceed accordingly. Follow the steps below to solve the problem:
- If the count of 1s and 0s are not equal in str1 and str2, then conversion is not possible.
- Traverse the string.
- Starting from the first character, compare each character one by one. For each different character at i, perform the following steps:
- Check if the current character of the string str1 is ‘0’ and curStr1Ones (stores the current count of 1’s of the string str1) is greater than 0. If found to be true, then swap the character with ‘1’ and decrement the value of curStr1Ones by 1.
- Check if the character of the string str1 is ‘0’ and curStr1Ones is equal to 0. If found to be true, then increment the value of the flag by 1 and break the loop.
- Check if the character of the string str1 is ‘1’ and the character of the string str2 is ‘0’. If found to be true, then swap the character of str1 with ‘0’ and increment the value of curStr1Ones by 1.
- Finally, print “Possible” if the flag is 0, otherwise print “Not Possible”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if it is possible to make // two binary strings equal by given operations void isBinaryStringsEqual(string str1, string str2) { // Stores count of 1's and 0's // of the string str1 int str1Zeros = 0, str1Ones = 0; // Stores count of 1's and 0's // of the string str2 int str2Zeros = 0, str2Ones = 0; int flag = 0; // Stores current count of 1's // present in the string str1 int curStr1Ones = 0; // Count the number of 1's and 0's // present in the strings str1 and str2 for ( int i = 0; i < str1.length(); i++) { if (str1[i] == '1' ) { str1Ones++; } else if (str1[i] == '0' ) { str1Zeros++; } if (str2[i] == '1' ) { str2Ones++; } else if (str2[i] == '0' ) { str2Zeros++; } } // If the number of 1's and 0's // are not same of the strings str1 // and str2 then print not possible if (str1Zeros != str2Zeros && str1Ones != str2Ones) { cout << "Not Possible" ; } else { // Traversing through the // strings str1 and str2 for ( int i = 0; i < str1.length(); i++) { // If the str1 character not // equals to str2 character if (str1[i] != str2[i]) { // Swaps 0 with 1 of the // string str1 if (str1[i] == '0' && curStr1Ones > 0) { str1[i] = '1' ; curStr1Ones--; } // Breaks the loop as the count // of 1's is zero. Hence, no swaps possible if (str1[i] == '0' && curStr1Ones == 0) { flag++; break ; } // Swaps 1 with 0 in the string str1 if (str1[i] == '1' && str2[i] == '0' ) { str1[i] = '0' ; curStr1Ones++; } } } if (flag == 0) { cout << "Possible" ; } // Print not possible else { cout << "Not Possible" ; } } } // Driver Code int main() { // Given Strings string str1 = "0110" ; string str2 = "0011" ; // Function Call isBinaryStringsEqual(str1, str2); return 0; } |
Java
// Java program for the above approach import java.io.*; class GFG { // Function to check if it is possible to make // two binary strings equal by given operations static void isBinaryStringsEqual(String str1, String str2) { // Stores count of 1's and 0's // of the string str1 int str1Zeros = 0 , str1Ones = 0 ; // Stores count of 1's and 0's // of the string str2 int str2Zeros = 0 , str2Ones = 0 ; int flag = 0 ; // Stores current count of 1's // present in the string str1 int curStr1Ones = 0 ; // Count the number of 1's and 0's // present in the strings str1 and str2 for ( int i = 0 ; i < str1.length(); i++) { if (str1.charAt(i) == '1' ) { str1Ones++; } else if (str1.charAt(i) == '0' ) { str1Zeros++; } if (str2.charAt(i) == '1' ) { str2Ones++; } else if (str2.charAt(i) == '0' ) { str2Zeros++; } } // If the number of 1's and 0's // are not same of the strings str1 // and str2 then print not possible if (str1Zeros != str2Zeros && str1Ones != str2Ones) { System.out.println( "Not Possible" ); } else { // Traversing through the // strings str1 and str2 for ( int i = 0 ; i < str1.length(); i++) { // If the str1 character not // equals to str2 character if (str1.charAt(i) != str2.charAt(i)) { // Swaps 0 with 1 of the // string str1 if (str1.charAt(i) == '0' && curStr1Ones > 0 ) { str1 = str1.substring( 0 , i) + '1' + str1.substring(i + 1 ); curStr1Ones--; } // Breaks the loop as the count // of 1's is zero. Hence, no swaps // possible if (str1.charAt(i) == '0' && curStr1Ones == 0 ) { flag++; break ; } // Swaps 1 with 0 in the string str1 if (str1.charAt(i) == '1' && str2.charAt(i) == '0' ) { str1 = str1.substring( 0 , i) + '0' + str1.substring(i+ 1 ); curStr1Ones++; } } } if (flag == 0 ) { System.out.println( "Possible" ); } // Print not possible else { System.out.println( "Not Possible" ); } } } // Driver Code public static void main(String[] args) { // Given Strings String str1 = "0110" ; String str2 = "0011" ; // Function Call isBinaryStringsEqual(str1, str2); } } // This code is contributed by dharanendralv23 |
Python3
# Python program for the above approach # Function to check if it is possible to make # two binary strings equal by given operations def isBinaryStringsEqual(list1, list2) : str1 = list (list1) str2 = list (list2) # Stores count of 1's and 0's # of the string str1 str1Zeros = 0 str1Ones = 0 # Stores count of 1's and 0's # of the string str2 str2Zeros = 0 str2Ones = 0 flag = 0 # Stores current count of 1's # present in the string str1 curStr1Ones = 0 # Count the number of 1's and 0's # present in the strings str1 and str2 for i in range ( len (str1)): if (str1[i] = = '1' ) : str1Ones + = 1 elif (str1[i] = = '0' ) : str1Zeros + = 1 if (str2[i] = = '1' ) : str2Ones + = 1 elif (str2[i] = = '0' ) : str2Zeros + = 1 # If the number of 1's and 0's # are not same of the strings str1 # and str2 then print not possible if (str1Zeros ! = str2Zeros and str1Ones ! = str2Ones) : print ( "Not Possible" ) else : # Traversing through the # strings str1 and str2 for i in range ( len (str1)): # If the str1 character not # equals to str2 character if (str1[i] ! = str2[i]) : # Swaps 0 with 1 of the # string str1 if (str1[i] = = '0' and curStr1Ones > 0 ) : str1[i] = '1' curStr1Ones - = 1 # Breaks the loop as the count # of 1's is zero. Hence, no swaps possible if (str1[i] = = '0' and curStr1Ones = = 0 ) : flag + = 1 break # Swaps 1 with 0 in the string str1 if (str1[i] = = '1' and str2[i] = = '0' ) : str1[i] = '0' curStr1Ones + = 1 if (flag = = 0 ) : print ( "Possible" ) # Print not possible else : print ( "Not Possible" ) # Driver Code # Given Strings str1 = "0110" str2 = "0011" # Function Call isBinaryStringsEqual(str1, str2) # This code is contributed by mohit kumar 29. |
C#
// C# program for the above approach using System; using System.Text; class GFG { // Function to check if it is possible to make // two binary strings equal by given operations static void isBinaryStringsEqual( string str1, string str2) { // Stores count of 1's and 0's // of the string str1 int str1Zeros = 0, str1Ones = 0; // Stores count of 1's and 0's // of the string str2 int str2Zeros = 0, str2Ones = 0; int flag = 0; // Stores current count of 1's // present in the string str1 int curStr1Ones = 0; // Count the number of 1's and 0's // present in the strings str1 and str2 for ( int i = 0; i < str1.Length; i++) { if (str1[i] == '1' ) { str1Ones++; } else if (str1[i] == '0' ) { str1Zeros++; } if (str2[i] == '1' ) { str2Ones++; } else if (str2[i] == '0' ) { str2Zeros++; } } // If the number of 1's and 0's // are not same of the strings str1 // and str2 then print not possible if (str1Zeros != str2Zeros && str1Ones != str2Ones) { Console.WriteLine( "Not Possible" ); } else { // Traversing through the // strings str1 and str2 for ( int i = 0; i < str1.Length; i++) { // If the str1 character not // equals to str2 character if (str1[i] != str2[i]) { // Swaps 0 with 1 of the // string str1 if (str1[i] == '0' && curStr1Ones > 0) { StringBuilder sb = new StringBuilder(str1); sb[i] = '1' ; str1 = sb.ToString(); curStr1Ones--; } // Breaks the loop as the count // of 1's is zero. Hence, no swaps // possible if (str1[i] == '0' && curStr1Ones == 0) { flag++; break ; } // Swaps 1 with 0 in the string str1 if (str1[i] == '1' && str2[i] == '0' ) { StringBuilder sb = new StringBuilder(str1); sb[i] = '0' ; str1 = sb.ToString(); curStr1Ones++; } } } if (flag == 0) { Console.WriteLine( "Possible" ); } // Print not possible else { Console.WriteLine( "Not Possible" ); } } } // Driver Code static public void Main() { // Given Strings string str1 = "0110" ; string str2 = "0011" ; // Function Call isBinaryStringsEqual(str1, str2); } } // This code is contributed by dharanendralv23 |
Javascript
<script> // JavaScript program for the above approach // Function to check if it is possible to make // two binary strings equal by given operations function isBinaryStringsEqual(list1, list2) { var str1 = list1.split( "" ); var str2 = list2.split( "" ); // Stores count of 1's and 0's // of the string str1 var str1Zeros = 0, str1Ones = 0; // Stores count of 1's and 0's // of the string str2 var str2Zeros = 0, str2Ones = 0; var flag = 0; // Stores current count of 1's // present in the string str1 var curStr1Ones = 0; // Count the number of 1's and 0's // present in the strings str1 and str2 for ( var i = 0; i < str1.length; i++) { if (str1[i] === "1" ) { str1Ones++; } else if (str1[i] === "0" ) { str1Zeros++; } if (str2[i] === "1" ) { str2Ones++; } else if (str2[i] === "0" ) { str2Zeros++; } } // If the number of 1's and 0's // are not same of the strings str1 // and str2 then print not possible if (str1Zeros !== str2Zeros && str1Ones !== str2Ones) { document.write( "Not Possible" ); } else { // Traversing through the // strings str1 and str2 for ( var i = 0; i < str1.length; i++) { // If the str1 character not // equals to str2 character if (str1[i] !== str2[i]) { // Swaps 0 with 1 of the // string str1 if (str1[i] === "0" && curStr1Ones > 0) { str1[i] = "1" ; curStr1Ones--; } // Breaks the loop as the count // of 1's is zero. Hence, no swaps // possible if (str1[i] === "0" && curStr1Ones === 0) { flag++; break ; } // Swaps 1 with 0 in the string str1 if (str1[i] === "1" && str2[i] === "0" ) { str1[i] = "0" ; curStr1Ones++; } } } if (flag === 0) { document.write( "Possible" ); } // Print not possible else { document.write( "Not Possible" ); } } } // Driver Code // Given Strings var str1 = "0110" ; var str2 = "0011" ; // Function Call isBinaryStringsEqual(str1, str2); </script> |
Possible
Time Complexity: O(|str1|)
Auxiliary Space: O(1)