Check if all substrings of length K of a Binary String has equal count of 0s and 1s

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 0s and 1s. If found to be true, print “Yes”. Otherwise, print “No”.

Examples:

Input: S = "101010", K = 2
Output: Yes
Explanation:
Since all the substrings of length 2 has equal number of 0s and 1s, the answer is Yes.

Input: S = "101011", K = 4
Output: No
Explanation:
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(N²) 
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++ 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 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 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# 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 

<script>
// javascript program for the
// above approach

// Function to check if the substring
// of length K has equal 0 and 1
function check(s, k)
{
  let n = s.length;
 
  // Traverse the String
  for (let i = 0; i < k; i++)
  {
    for (let j = i; j < n; j += k)
    {
      // Check if every K-th character
      // is the same or not
      if (s[i] != s[j])
        return false;
    }
  }
  let c = 0;
 
  // Traverse subString of length K
  for (let 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

     let s = "101010";
  let k = 2;
 
  if (check(s, k))
    document.write("Yes" + "<br/>");
  else
    document.write("No");

// This code is contributed by target_2.
</script>

Yes

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