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Generate a Binary String without any consecutive 0’s and at most K consecutive 1’s
  • Difficulty Level : Medium
  • Last Updated : 10 May, 2021

Given two integers N and M, the task is to construct a binary string with the following conditions : 

  • The Binary String consists of N 0’s and M 1’s
  • The Binary String has at most K consecutive 1’s.
  • The Binary String does not contain any adjacent 0’s.

If it is not possible to construct such a binary string, then print -1.

Examples: 
 

Input: N = 5, M = 9, K = 2 
Output: 01101101101101 
Explanation: 
The string “01101101101101” satisfies the following conditions: 
 

  • No consecutive 0’s are present.
  • No more than K(= 2) consecutive 1’s are present.

Input: N = 4, M = 18, K = 4 
Output: 1101111011110111101111 
 



 

 

Approach:  

To construct a binary string satisfying the given properties, observe the following:

  • For no two ‘0‘s to be consecutive, there should be at least a ‘1‘ placed between them.
  • Therefore, for N number of ‘0‘s, there should be at least N-11‘s present for a string of required type to be generated.
  • Since no more than K consecutive ‘1‘s can be placed together, for N 0’s, there can be a maximum (N+1) * K ‘1‘s possible.
  • Therefore, the number of ‘1‘s should lie within the range: 
     

 
 

N – 1 ? M ? (N + 1) * K 
 

 

  • If the given values N and M do not satisfy the above condition, then print -1.
  • Otherwise, follow the steps below to solve the problem:
    • Append ‘0‘s to the final string.
    • Insert ‘1‘ in between each pair of 0′s. Subtract N – 1 from M, as N – 11‘s have already been placed.
    • For the remaining ‘1‘s, place min(K – 1, M)1‘s alongside each already placed ‘1‘s, to ensure that no more than K ‘1’s are placed together.
    • For any remaining ‘1‘s, append them to the beginning and end of the final string.
  • Finally, print the string generated.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to construct the binary string
string ConstructBinaryString(int N, int M,
                             int K)
{
    // Conditions when string construction
    // is not possible
    if (M < (N - 1) || M > K * (N + 1))
        return "-1";
 
    string ans = "";
 
    // Stores maximum 1's that
    // can be placed in between
    int l = min(K, M / (N - 1));
    int temp = N;
    while (temp--) {
        // Place 0's
        ans += '0';
 
        if (temp == 0)
            break;
 
        // Place 1's in between
        for (int i = 0; i < l; i++) {
            ans += '1';
        }
    }
 
    // Count remaining M's
    M -= (N - 1) * l;
 
    if (M == 0)
        return ans;
 
    l = min(M, K);
    // Place 1's at the end
    for (int i = 0; i < l; i++)
        ans += '1';
 
    M -= l;
    // Place 1's at the beginning
    while (M > 0) {
        ans = '1' + ans;
        M--;
    }
 
    // Return the final string
    return ans;
}
 
// Driver Code
int main()
{
    int N = 5, M = 9, K = 2;
 
    cout << ConstructBinaryString(N, M, K);
}

Java




// Java implementation of
// the above approach
class GFG{
     
// Function to construct the binary string
static String ConstructBinaryString(int N, int M,
                                    int K)
{
     
    // Conditions when string construction
    // is not possible
    if (M < (N - 1) || M > K * (N + 1))
        return "-1";
 
    String ans = "";
 
    // Stores maximum 1's that
    // can be placed in between
    int l = Math.min(K, M / (N - 1));
    int temp = N;
     
    while (temp != 0)
    {
        temp--;
         
        // Place 0's
        ans += '0';
 
        if (temp == 0)
            break;
 
        // Place 1's in between
        for(int i = 0; i < l; i++)
        {
            ans += '1';
        }
    }
 
    // Count remaining M's
    M -= (N - 1) * l;
 
    if (M == 0)
        return ans;
 
    l = Math.min(M, K);
     
    // Place 1's at the end
    for(int i = 0; i < l; i++)
        ans += '1';
 
    M -= l;
     
    // Place 1's at the beginning
    while (M > 0)
    {
        ans = '1' + ans;
        M--;
    }
 
    // Return the final string
    return ans;
}
 
// Driver code   
public static void main(String[] args)
{
    int N = 5, M = 9, K = 2;
     
    System.out.println(ConstructBinaryString(N, M, K));
}
}
 
// This code is contributed by rutvik_56

Python3




# Python3 implementation of
# the above approach
 
# Function to construct the binary string
def ConstructBinaryString(N, M, K):
 
    # Conditions when string construction
    # is not possible
    if(M < (N - 1) or M > K * (N + 1)):
        return '-1'
 
    ans = ""
 
    # Stores maximum 1's that
    # can be placed in between
    l = min(K, M // (N - 1))
    temp = N
     
    while(temp):
        temp -= 1
 
        # Place 0's
        ans += '0'
 
        if(temp == 0):
            break
 
        # Place 1's in between
        for i in range(l):
            ans += '1'
 
    # Count remaining M's
    M -= (N - 1) * l
 
    if(M == 0):
        return ans
 
    l = min(M, K)
     
    # Place 1's at the end
    for i in range(l):
        ans += '1'
 
    M -= l
     
    # Place 1's at the beginning
    while(M > 0):
        ans = '1' + ans
        M -= 1
 
    # Return the final string
    return ans
 
# Driver Code
if __name__ == '__main__':
 
    N = 5
    M = 9
    K = 2
     
    print(ConstructBinaryString(N, M , K))
 
# This code is contributed by Shivam Singh

C#




// C# implementation of
// the above approach
using System;
class GFG{
      
// Function to construct the binary string
static String ConstructBinaryString(int N, int M,
                                    int K)
{
      
    // Conditions when string construction
    // is not possible
    if (M < (N - 1) || M > K * (N + 1))
        return "-1";
  
    string ans = "";
  
    // Stores maximum 1's that
    // can be placed in between
    int l = Math.Min(K, M / (N - 1));
    int temp = N;
      
    while (temp != 0)
    {
        temp--;
          
        // Place 0's
        ans += '0';
  
        if (temp == 0)
            break;
  
        // Place 1's in between
        for(int i = 0; i < l; i++)
        {
            ans += '1';
        }
    }
  
    // Count remaining M's
    M -= (N - 1) * l;
  
    if (M == 0)
        return ans;
  
    l = Math.Min(M, K);
      
    // Place 1's at the end
    for(int i = 0; i < l; i++)
        ans += '1';
  
    M -= l;
      
    // Place 1's at the beginning
    while (M > 0)
    {
        ans = '1' + ans;
        M--;
    }
  
    // Return the final string
    return ans;
}
  
// Driver code   
public static void Main(string[] args)
{
    int N = 5, M = 9, K = 2;
      
    Console.Write(ConstructBinaryString(N, M, K));
}
}
  
// This code is contributed by Ritik Bansal

Javascript




<script>
// JavaScript program for the above approach
 
// Function to construct the binary string
function ConstructBinaryString(N, M, K)
{
      
    // Conditions when string construction
    // is not possible
    if (M < (N - 1) || M > K * (N + 1))
        return "-1";
  
    let ans = "";
  
    // Stores maximum 1's that
    // can be placed in between
    let l = Math.min(K, M / (N - 1));
    let temp = N;
      
    while (temp != 0)
    {
        temp--;
          
        // Place 0's
        ans += '0';
  
        if (temp == 0)
            break;
  
        // Place 1's in between
        for(let i = 0; i < l; i++)
        {
            ans += '1';
        }
    }
  
    // Count remaining M's
    M -= (N - 1) * l;
  
    if (M == 0)
        return ans;
  
    l = Math.min(M, K);
      
    // Place 1's at the end
    for(let i = 0; i < l; i++)
        ans += '1';
  
    M -= l;
      
    // Place 1's at the beginning
    while (M > 0)
    {
        ans = '1' + ans;
        M--;
    }
  
    // Return the final string
    return ans;
}
 
// Driver Code   
     
        let N = 5, M = 9, K = 2;
      
    document.write(ConstructBinaryString(N, M, K));
                     
</script>
Output: 
01101101101101

 

Time Complexity: O(N+M)
Auxiliary Space: O(N+M)

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