Print n 0s and m 1s such that no two 0s and no three 1s are together

Given two integers n and m where n is the number of 0s and m is the number of 1s. The task is to print all the 0s and 1s in a single row such that no two 0s are together and no three 1s are together. If it’s not possible to arrange 0s and 1s according to the condition then print -1.

Examples:

Input: n = 1, m = 2
Output: 011

Input: n = 4, m = 8
Output: 110110110101

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We have answers only when, ( (n – 1) ≤ m and m ≤ 2 * (n + 1).

• If (m == n – 1) then print the pattern 010101… starting from 0 until all the 0s and 1s have been used.
• If (m > n) and m ≤ 2 * (n + 1) then print the pattern 110110110… until there is escessive 1s and change to pattern 0101010… when m becomes equal to n – 1.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to print the required pattern void printPattern(int n, int m) {        // When condition fails     if (m > 2 * (n + 1) || m < n - 1) {         cout << "-1";     }        // When m = n - 1     else if (abs(n - m) <= 1) {         while (n > 0 && m > 0) {             cout << "01";             n--;             m--;         }         if (n != 0) {             cout << "0";         }         if (m != 0) {             cout << "1";         }     }     else {         while (m - n > 1 && n > 0) {             cout << "110";             m = m - 2;             n = n - 1;         }         while (n > 0) {             cout << "10";             n--;             m--;         }         while (m > 0) {             cout << "1";             m--;         }     } }    // Driver program int main() {     int n = 4, m = 8;     printPattern(n, m);     return 0; }

Java

 // Java implementation of the above approach  class GFG  {      // Function to print the required pattern      static void printPattern(int n, int m)      {          // When condition fails          if (m > 2 * (n + 1) || m < n - 1)          {              System.out.print("-1");          }                     // When m = n - 1          else if (Math.abs(n - m) <= 1)          {              while (n > 0 && m > 0)              {                  System.out.print("01");                  n--;                  m--;                                 }              if (n != 0)              {                  System.out.print("0");              }              if (m != 0)              {                  System.out.print("1");              }         }          else         {              while (m - n > 1 && n > 0)              {                  System.out.print("110");                  m = m - 2;                  n = n - 1;              }              while (n > 0)              {                  System.out.print("10");                  n--;                  m--;              }              while (m > 0)              {                  System.out.print("1");                  m--;              }          }      }         // Driver code      public static void main(String []args)      {          int n = 4, m = 8;          printPattern(n, m);      }  }     // This code is contributed by Ita_c.

Python3

 # Python 3 implementation of the approach    # Function to print the required pattern def printPattern(n, m):            # When condition fails     if (m > 2 * (n + 1) or m < n - 1):         print("-1", end = "")        # When m = n - 1     elif (abs(n - m) <= 1):         while (n > 0 and m > 0):             print("01", end = "");             n -= 1             m -= 1                    if (n != 0):             print("0", end = "")         if (m != 0):             print("1", end = "")     else:         while (m - n > 1 and n > 0):             print("110", end = "")             m = m - 2             n = n - 1                    while (n > 0):             print("10", end = "")             n -= 1             m -= 1                    while (m > 0):             print("1", end = "")             m -= 1        # Driver Code if __name__ == '__main__':     n = 4     m = 8     printPattern(n, m)    # This code is contributed by # Surendra_Gangwar

C#

 // C# implementation of the above approach  using System;    class GFG  {      // Function to print the required pattern      static void printPattern(int n, int m)      {          // When condition fails          if (m > 2 * (n + 1) || m < n - 1)          {              Console.Write("-1");          }          // When m = n - 1          else if (Math.Abs(n - m) <= 1)          {              while (n > 0 && m > 0)              {                  Console.Write("01");                  n--;                  m--;                                 }              if (n != 0)              {                  Console.Write("0");              }              if (m != 0)              {                  Console.Write("1");              }         }          else          {              while (m - n > 1 && n > 0)               {                  Console.Write("110");                  m = m - 2;                  n = n - 1;              }              while (n > 0)              {                  Console.Write("10");                  n--;                  m--;              }              while (m > 0)              {                  Console.Write("1");                  m--;              }          }      }         // Driver code      public static void Main()      {          int n = 4, m = 8;          printPattern(n, m);      }  }     // This code is contributed by Ryuga

PHP

 2 * (\$n + 1) || \$m < \$n - 1)      {          echo("-1");      }             // When m = n - 1      else if (abs(\$n - \$m) <= 1)      {          while (\$n > 0 && \$m > 0)          {              System.out.print("01");              \$n--;              \$m--;                         }          if (\$n != 0)          {              echo("0");          }          if (\$m != 0)          {              echo("1");          }     }      else     {          while (\$m - \$n > 1 && \$n > 0)          {              echo("110");              \$m = \$m - 2;              \$n = \$n - 1;          }          while (\$n > 0)          {              echo("10");              \$n--;              \$m--;          }          while (\$m > 0)          {              echo("1");              \$m--;          }      }  }     // Driver code  \$n = 4; \$m = 8;      printPattern(\$n, \$m);     // This code is contributed by // Mukul Singh. ?>

Output:

110110110101

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