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:
- 000 -> “ABCD”
- 001 -> “ABC D”
- 010 -> “AB CD”
- 011 -> “AB C D”
- 100 -> “A BCD”
- 101 -> “A BC D”
- 110 -> “A B CD”
- 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 characterslong 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 Codeint 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 characterslong 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 Codeint 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 charactersstatic 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 Codepublic 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 charactersdef 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 Codeif __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 charactersstatic 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 Codepublic static void Main(String[] args){ string S = "ABCD"; Console.WriteLine(countNumberOfStrings(S));}}// This code is contributed by AnkThon |
8
Time Complexity: O(log (len – 1)), where len represents length of the given string.
Auxiliary Space: O(1)
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