Count all possible strings that can be generated by placing spaces

Given a string S, the task is to count all possible strings that can be generated by placing spaces between any pair of adjacent characters of the string.

Examples:

Input: S = “AB”
Output: 2
Explanation: All possible strings are { “A B”, “AB”}.

Input: S = “ABC”
Output: 4
Explanation: All possible strings are {“A BC”, “AB C”, “A B C”, “ABC”}

Approach: The problem can be solved by assuming the spaces between adjacent pair of characters of the string as binary bits. Generally, if the length of the string is L, then there L – 1 places to fill by spaces.

Illustration:

S = “ABCD” 
Possible places for spaces are: 

• Between “A” and “B”
• Between “B” and “C”
• Between “C” and “D”

Length of the string = 4 
Possible spaces for spaces = 3 = 4 – 1
Assuming each place to be a binary bit, the total number of possible combinations are: 

1. 000 -> “ABCD”
2. 001 -> “ABC D”
3. 010 -> “AB CD”
4. 011 -> “AB C D”
5. 100 -> “A BCD”
6. 101 -> “A BC D”
7. 110 -> “A B CD”
8. 111 -> “A B C D”

Hence, 8 possible strings can be obtained for a string of length 4.
Therefore, total count of strings = 2 ^{L – 1}

Below is the implementation of the above idea: 

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of strings
// that can be generated by placing spaces
// between pair of adjacent characters

long long int countNumberOfStrings(string s)
{

    // Length of the string
    int length = s.length();

    // Count of positions for spaces
    int n = length - 1;

    // Count of possible strings
    long long int count = pow(2, n);

    return count;
}

// Driver Code
int main()
{
    string S = "ABCD";
    cout << countNumberOfStrings(S);

    return 0;
}

C

// C program to implement
// the above approach
#include <math.h>
#include <stdio.h>
#include <string.h>

// Function to count the number of strings
// that can be generated by placing spaces
// between pair of adjacent characters
long long int countNumberOfStrings(char* s)
{

    // Length of the string
    int length = strlen(s);

    // Count of positions for spaces
    int n = length - 1;

    // Count of possible strings
    long long int count = pow(2, n);

    return count;
}

// Driver Code
int main()
{
    char S[] = "ABCD";
    printf("%lld", countNumberOfStrings(S));

    return 0;
}

// This code is contributed by single__loop

Java

// Java program to implement
// the above approach
import java.io.*;

class GFG{

// Function to count the number of strings
// that can be generated by placing spaces
// between pair of adjacent characters
static long countNumberOfStrings(String s)
{

    // Count of positions for spaces
    int n = s.length() - 1;

    // Count of possible strings
    long count = (long)(Math.pow(2, n));

    return count;
}

// Driver Code
public static void main(String[] args)
{
    String S = "ABCD";

    System.out.println(countNumberOfStrings(S));
}
}

// This code is contributed by single__loop

Python3

# Python3 program to implement
# the above approach

# Function to count the number of strings
# that can be generated by placing spaces
# between pair of adjacent characters
def countNumberOfStrings(s):

    # Length of the string
    length = len(s)

    # Count of positions for spaces
    n = length - 1

    # Count of possible strings
    count = 2 ** n

    return count

# Driver Code
if __name__ == "__main__" :

    S = "ABCD"

    print(countNumberOfStrings(S))

# This code is contributed by AnkThon

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to count the number of strings
// that can be generated by placing spaces
// between pair of adjacent characters
static long countNumberOfStrings(String s)
{

    // Count of positions for spaces
    int n = s.Length - 1;

    // Count of possible strings
    long count = (long)(Math.Pow(2, n));

    return count;
}

// Driver Code
public static void Main(String[] args)
{
    string S = "ABCD";

    Console.WriteLine(countNumberOfStrings(S));
}
}

// This code is contributed by AnkThon

Javascript

<script>

// JavaScript program for above approach

// Function to count the number of strings
// that can be generated by placing spaces
// between pair of adjacent characters
function countNumberOfStrings(s)
{

    // Count of positions for spaces
    let n = s.length - 1;

    // Count of possible strings
    let count = (Math.pow(2, n));

    return count;
}

// Driver Code
     let S = "ABCD";
    document.write(countNumberOfStrings(S));

  // This code is contributed by avijitmondal1998.
</script>

Output:

8

Time Complexity: O(log (len – 1)), where len represents length of the given string.
Auxiliary Space: O(1)