Given a binary string **S **of length **N** and an even integer **K**, the task is to check if all substrings of length **K** contains an equal number of **0**s and **1**s. If found to be true, print “Yes”. Otherwise, print “No”.

**Examples:**

Input:S = “101010”, K = 2Output:YesExplanation:

Since all the substrings of length 2 has equal number of 0s and 1s, the answer is Yes.

Input:S = “101011”, K = 4Output:NoExplanation:

Since substring “1011” has unequal count of 0s and 1s, the answer is No..

**Naive Approach: **The simplest approach to solve the problem is to generate all substrings of length K and check it if it contains an equal count of 1s and 0s or not.

**Time Complexity: **O(N^{2})**Auxiliary Space:** O(1)

**Efficient Approach: **The main observation for optimizing the above approach is, for the string S to have an equal count of 0 and 1 in substrings of length **K**, **S[i]** must be equal to **S[i + k]**. Follow the steps below to solve the problem:

- Traverse the string and for every i
^{th}character, check if**S[i] = S[j]**where**(i == (j % k))** - If found to be false at any instant, then print “No”.
- Otherwise, check the first substring of length K and if it contains an equal count of 1s and 0s or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to check if the substring` `// of length K has equal 0 and 1` `int` `check(string& s, ` `int` `k)` `{` ` ` `int` `n = s.size();` ` ` `// Traverse the string` ` ` `for` `(` `int` `i = 0; i < k; i++) {` ` ` `for` `(` `int` `j = i; j < n; j += k) {` ` ` `// Check if every K-th character` ` ` `// is the same or not` ` ` `if` `(s[i] != s[j])` ` ` `return` `false` `;` ` ` `}` ` ` `}` ` ` `int` `c = 0;` ` ` `// Traverse substring of length K` ` ` `for` `(` `int` `i = 0; i < k; i++) {` ` ` `// If current character is 0` ` ` `if` `(s[i] == ` `'0'` `)` ` ` `// Increment count` ` ` `c++;` ` ` `// Otherwise` ` ` `else` ` ` `// Decrement count` ` ` `c--;` ` ` `}` ` ` `// Check for equal 0s and 1s` ` ` `if` `(c == 0)` ` ` `return` `true` `;` ` ` `else` ` ` `return` `false` `;` `}` `// Driver code` `int` `main()` `{` ` ` `string s = ` `"101010"` `;` ` ` `int` `k = 2;` ` ` `if` `(check(s, k))` ` ` `cout << ` `"Yes"` `<< endl;` ` ` `else` ` ` `cout << ` `"No"` `<< endl;` ` ` `return` `0;` `}` |

## Java

`// Java program for` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to check if the substring` `// of length K has equal 0 and 1` `static` `boolean` `check(String s, ` `int` `k)` `{` ` ` `int` `n = s.length();` ` ` `// Traverse the String` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++)` ` ` `{` ` ` `for` `(` `int` `j = i; j < n; j += k)` ` ` `{` ` ` `// Check if every K-th character` ` ` `// is the same or not` ` ` `if` `(s.charAt(i) != s.charAt(j))` ` ` `return` `false` `;` ` ` `}` ` ` `}` ` ` `int` `c = ` `0` `;` ` ` `// Traverse subString of length K` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++)` ` ` `{` ` ` `// If current character is 0` ` ` `if` `(s.charAt(i) == ` `'0'` `)` ` ` `// Increment count` ` ` `c++;` ` ` `// Otherwise` ` ` `else` ` ` `// Decrement count` ` ` `c--;` ` ` `}` ` ` `// Check for equal 0s and 1s` ` ` `if` `(c == ` `0` `)` ` ` `return` `true` `;` ` ` `else` ` ` `return` `false` `;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `String s = ` `"101010"` `;` ` ` `int` `k = ` `2` `;` ` ` `if` `(check(s, k))` ` ` `System.out.print(` `"Yes"` `+ ` `"\n"` `);` ` ` `else` ` ` `System.out.print(` `"No"` `+ ` `"\n"` `);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 program for the above approach` ` ` `# Function to check if the substring` `# of length K has equal 0 and 1` `def` `check(s, k):` ` ` ` ` `n ` `=` `len` `(s)` ` ` ` ` `# Traverse the string` ` ` `for` `i ` `in` `range` `(k):` ` ` `for` `j ` `in` `range` `(i, n, k):` ` ` ` ` `# Check if every K-th character` ` ` `# is the same or not` ` ` `if` `(s[i] !` `=` `s[j]):` ` ` `return` `False` ` ` ` ` `c ` `=` `0` ` ` ` ` `# Traverse substring of length K` ` ` `for` `i ` `in` `range` `(k):` ` ` ` ` `# If current character is 0` ` ` `if` `(s[i] ` `=` `=` `'0'` `):` ` ` ` ` `# Increment count` ` ` `c ` `+` `=` `1` ` ` ` ` `# Otherwise` ` ` `else` `:` ` ` ` ` `# Decrement count` ` ` `c ` `-` `=` `1` ` ` ` ` `# Check for equal 0s and 1s` ` ` `if` `(c ` `=` `=` `0` `):` ` ` `return` `True` ` ` `else` `:` ` ` `return` `False` `# Driver code` `s ` `=` `"101010"` `k ` `=` `2` ` ` `if` `(check(s, k) !` `=` `0` `):` ` ` `print` `(` `"Yes"` `)` `else` `:` ` ` `print` `(` `"No"` `)` ` ` `# This code is contributed by sanjoy_62` |

## C#

`// C# program for` `// the above approach` `using` `System;` `class` `GFG{` `// Function to check if the substring` `// of length K has equal 0 and 1` `static` `bool` `check(String s, ` `int` `k)` `{` ` ` `int` `n = s.Length;` ` ` `// Traverse the String` ` ` `for` `(` `int` `i = 0; i < k; i++)` ` ` `{` ` ` `for` `(` `int` `j = i; j < n; j += k)` ` ` `{` ` ` `// Check if every K-th character` ` ` `// is the same or not` ` ` `if` `(s[i] != s[j])` ` ` `return` `false` `;` ` ` `}` ` ` `}` ` ` `int` `c = 0;` ` ` `// Traverse subString of length K` ` ` `for` `(` `int` `i = 0; i < k; i++)` ` ` `{` ` ` `// If current character is 0` ` ` `if` `(s[i] == ` `'0'` `)` ` ` `// Increment count` ` ` `c++;` ` ` `// Otherwise` ` ` `else` ` ` `// Decrement count` ` ` `c--;` ` ` `}` ` ` `// Check for equal 0s and 1s` ` ` `if` `(c == 0)` ` ` `return` `true` `;` ` ` `else` ` ` `return` `false` `;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `String s = ` `"101010"` `;` ` ` `int` `k = 2;` ` ` `if` `(check(s, k))` ` ` `Console.Write(` `"Yes"` `+ ` `"\n"` `);` ` ` `else` ` ` `Console.Write(` `"No"` `+ ` `"\n"` `);` `}` `}` `// This code is contributed by Rajput-Ji` |

**Output:**

Yes

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.