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-1 ‘1‘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 – 1 ‘1‘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++ 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 implementation of // the above approach import java.io.*;
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 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# 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 |
<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> |
01101101101101
Time Complexity: O(N+M)
Auxiliary Space: O(N+M)