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
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 excessive 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 <bits/stdc++.h>using namespace std;// Function to print the required patternvoid 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 programint 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 patterndef 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 Codeif __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
<?php// PHP implementation of the above approach // Function to print the required pattern function printPattern($n, $m) { // When condition fails if ($m > 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.?> |
Javascript
<script>// JavaScript implementation of the above approach // Function to print the required pattern function printPattern( n, m){ // When condition fails if (m > 2 * (n + 1) || m < n - 1) { document.write("-1"); } // When m = n - 1 else if (Math.abs(n - m) <= 1) { while (n > 0 && m > 0) { document.write("01"); n--; m--; } if (n != 0) { document.write("0"); } if (m != 0) { document.write("1"); } } else { while (m - n > 1 && n > 0) { document.write("110"); m = m - 2; n = n - 1; } while (n > 0) { document.write("10"); n--; m--; } while (m > 0) { document.write("1"); m--; } }}var n = 4, m = 8; printPattern(n, m);</script> |
Output
110110110101
Time complexity : O(m)
Space complexity : O(1)
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