Given two binary strings **S1 **and **S2** of length **N** (*1 ≤ N ≤ 10 ^{5}*), the task is to check if it is possible to convert the string

**S1**to

**S2**by performing the following operations any number of times:

- Select any two indices
**i**and**j**(*1 ≤ i < j ≤ N*) such that**S1[i]**is**‘0’**and**S1[j]**is**‘1’**. - Swap
**S1[i]**with**S1[j]**.

**Examples:**

Input:S1 =”100111″, S2 = “111010”Output: YESExplanation:Swap S[2] and S[3] with S[4] and S[6] respectively.

Input:S1 = “110100”, S2 = “010101”Output:NO

**Approach:** Follow the steps below to solve the problem:

- Check if the number of occurrences of character
**‘0’**and**‘1’**in both the strings are equal. If not found to be true, then it is impossible to transform string**S1**into**S2**. - If the number of characters are equal, then we move to the next step. By the given condition it is possible to move the ‘0’ only in
**forward direction**by swapping with letter ‘1’ in string**S1**. - Hence, iterate characters of both the strings and count the number of occurrences of
**‘0’**in both strings. If at any point the count of characters of ‘0’ in string**S2**becomes strictly greater than count of occurrences in string**S1**, terminate the loop and print**“NO”**. - If both the strings iterated successfully, print
**“YES”**.

Below is the implementation of above approach:

## C++

`// C++ Program to implement` `// of above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if a string` `// s1 can be converted into s2` `void` `check(string s1, string s2)` `{` ` ` `// Count of '0' in strings in s1 and s2` ` ` `int` `s1_0 = 0, s2_0 = 0;` ` ` `// Iterate both the strings and` ` ` `// count the number of occurrences of` ` ` `for` `(` `int` `i = 0; i < s1.size(); i++) {` ` ` `if` `(s1[i] == ` `'0'` `) {` ` ` `s1_0++;` ` ` `}` ` ` `if` `(s2[i] == ` `'0'` `) {` ` ` `s2_0++;` ` ` `}` ` ` `}` ` ` `// Count is not equal` ` ` `if` `(s1_0 != s2_0) {` ` ` `cout << ` `"NO"` `<< endl;` ` ` `return` `;` ` ` `}` ` ` `else` `{` ` ` `int` `Count1 = 0, Count2 = 0;` ` ` `// Iterating over both the` ` ` `// arrays and count the` ` ` `// number of occurrences of '0'` ` ` `for` `(` `int` `i = 0; i < s1.size(); i++) {` ` ` `if` `(s1[i] == ` `'0'` `) {` ` ` `Count1++;` ` ` `}` ` ` `if` `(s2[i] == ` `'0'` `) {` ` ` `Count2++;` ` ` `}` ` ` `// If the count of occurrences` ` ` `// of '0' in S2 exceeds that in S1` ` ` `if` `(Count1 < Count2) {` ` ` `cout << ` `"NO"` `<< endl;` ` ` `return` `;` ` ` `}` ` ` `}` ` ` `cout << ` `"YES"` `<< endl;` ` ` `}` `}` `// Driver program` `int` `main()` `{` ` ` `string s1 = ` `"100111"` `;` ` ` `string s2 = ` `"111010"` `;` ` ` `check(s1, s2);` ` ` `s1 = ` `"110100"` `;` ` ` `s2 = ` `"010101"` `;` ` ` `check(s1, s2);` ` ` `return` `0;` `}` |

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## Java

`// Java program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG` `{` ` ` `// Function to check if a string ` `// s1 can be converted into s2 ` `static` `void` `check(String s1, String s2) ` `{ ` ` ` ` ` `// Count of '0' in strings in s1 and s2 ` ` ` `int` `s1_0 = ` `0` `, s2_0 = ` `0` `; ` ` ` ` ` `// Iterate both the strings and ` ` ` `// count the number of occurrences of ` ` ` `for` `(` `int` `i = ` `0` `; i < s1.length(); i++) ` ` ` `{ ` ` ` `if` `(s1.charAt(i) == ` `'0'` `) ` ` ` `{` ` ` `s1_0++; ` ` ` `} ` ` ` ` ` `if` `(s2.charAt(i) == ` `'0'` `) ` ` ` `{` ` ` `s2_0++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Count is not equal ` ` ` `if` `(s1_0 != s2_0) ` ` ` `{` ` ` `System.out.println(` `"NO"` `); ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `else` ` ` `{ ` ` ` `int` `Count1 = ` `0` `, Count2 = ` `0` `; ` ` ` ` ` `// Iterating over both the ` ` ` `// arrays and count the ` ` ` `// number of occurrences of '0' ` ` ` `for` `(` `int` `i = ` `0` `; i < s1.length(); i++)` ` ` `{ ` ` ` `if` `(s1.charAt(i) == ` `'0'` `)` ` ` `{` ` ` `Count1++; ` ` ` `} ` ` ` ` ` `if` `(s2.charAt(i) == ` `'0'` `)` ` ` `{` ` ` `Count2++; ` ` ` `} ` ` ` ` ` `// If the count of occurrences ` ` ` `// of '0' in S2 exceeds that in S1 ` ` ` `if` `(Count1 < Count2) ` ` ` `{` ` ` `System.out.println(` `"NO"` `); ` ` ` `return` `; ` ` ` `} ` ` ` `} ` ` ` `System.out.println(` `"YES"` `); ` ` ` `} ` `} ` ` ` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `String s1 = ` `"100111"` `; ` ` ` `String s2 = ` `"111010"` `; ` ` ` `check(s1, s2); ` ` ` `s1 = ` `"110100"` `; ` ` ` `s2 = ` `"010101"` `; ` ` ` `check(s1, s2);` `}` `}` `// This code is contributed by code_hunt.` |

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## Python3

`# Python3 program to implement` `# of above approach` `# Function to check if a string` `# s1 can be converted into s2` `def` `check(s1, s2):` ` ` `# Count of '0' in strings in s1 and s2` ` ` `s1_0 ` `=` `0` ` ` `s2_0 ` `=` `0` ` ` `# Iterate both the strings and` ` ` `# count the number of occurrences of` ` ` `for` `i ` `in` `range` `(` `len` `(s1)):` ` ` `if` `(s1[i] ` `=` `=` `'0'` `):` ` ` `s1_0 ` `+` `=` `1` ` ` ` ` `if` `(s2[i] ` `=` `=` `'0'` `):` ` ` `s2_0 ` `+` `=` `1` ` ` `# Count is not equal` ` ` `if` `(s1_0 !` `=` `s2_0):` ` ` `print` `(` `"NO"` `)` ` ` `return` ` ` `else` `:` ` ` `Count1 ` `=` `0` ` ` `Count2 ` `=` `0` `;` ` ` `# Iterating over both the` ` ` `# arrays and count the` ` ` `# number of occurrences of '0'` ` ` `for` `i ` `in` `range` `(` `len` `(s1)):` ` ` `if` `(s1[i] ` `=` `=` `'0'` `):` ` ` `Count1 ` `+` `=` `1` ` ` `if` `(s2[i] ` `=` `=` `'0'` `):` ` ` `Count2 ` `+` `=` `1` ` ` `# If the count of occurrences` ` ` `# of '0' in S2 exceeds that in S1` ` ` `if` `(Count1 < Count2):` ` ` `print` `(` `"NO"` `)` ` ` `return` ` ` ` ` `print` `(` `"YES"` `)` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `s1 ` `=` `"100111"` ` ` `s2 ` `=` `"111010"` ` ` `check(s1, s2)` ` ` `s1 ` `=` `"110100"` ` ` `s2 ` `=` `"010101"` ` ` `check(s1, s2)` `# This code is contributed by chitranayal` |

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## C#

`// C# program to implement ` `// of above approach ` `using` `System;` `class` `GFG{` ` ` `// Function to check if a string ` `// s1 can be converted into s2 ` `static` `void` `check(` `string` `s1, ` `string` `s2) ` `{ ` ` ` ` ` `// Count of '0' in strings in s1 and s2 ` ` ` `int` `s1_0 = 0, s2_0 = 0; ` ` ` ` ` `// Iterate both the strings and ` ` ` `// count the number of occurrences of ` ` ` `for` `(` `int` `i = 0; i < s1.Length; i++) ` ` ` `{ ` ` ` `if` `(s1[i] == ` `'0'` `) ` ` ` `{` ` ` `s1_0++; ` ` ` `} ` ` ` ` ` `if` `(s2[i] == ` `'0'` `) ` ` ` `{` ` ` `s2_0++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Count is not equal ` ` ` `if` `(s1_0 != s2_0) ` ` ` `{` ` ` `Console.WriteLine(` `"NO"` `); ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `else` ` ` `{ ` ` ` `int` `Count1 = 0, Count2 = 0; ` ` ` ` ` `// Iterating over both the ` ` ` `// arrays and count the ` ` ` `// number of occurrences of '0' ` ` ` `for` `(` `int` `i = 0; i < s1.Length; i++)` ` ` `{ ` ` ` `if` `(s1[i] == ` `'0'` `)` ` ` `{` ` ` `Count1++; ` ` ` `} ` ` ` ` ` `if` `(s2[i] == ` `'0'` `)` ` ` `{` ` ` `Count2++; ` ` ` `} ` ` ` ` ` `// If the count of occurrences ` ` ` `// of '0' in S2 exceeds that in S1 ` ` ` `if` `(Count1 < Count2) ` ` ` `{` ` ` `Console.WriteLine(` `"NO"` `); ` ` ` `return` `; ` ` ` `} ` ` ` `} ` ` ` `Console.WriteLine(` `"YES"` `); ` ` ` `} ` `} ` `// Driver code` `static` `void` `Main() ` `{` ` ` `string` `s1 = ` `"100111"` `; ` ` ` `string` `s2 = ` `"111010"` `; ` ` ` ` ` `check(s1, s2); ` ` ` ` ` `s1 = ` `"110100"` `; ` ` ` `s2 = ` `"010101"` `; ` ` ` ` ` `check(s1, s2);` `}` `}` `// This code is contributed by divyesh072019` |

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**Output:**

YES NO

**Time Complexity: **O(N)**Auxiliary Space**: O(1)

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