String str is given which contains lowercase English letters and spaces. It may contain multiple spaces. Get the first letter of every word and return the result as a string. The result should not contain any space.
Examples:
Input: str = "neveropen for neveropen" Output: gfg Input: str = "happy coding" Output: hc
Source: https://www.geeksforgeeks.org/amazon-interview-set-8-2/
The idea is to traverse each character of string str and maintain a boolean variable, which was initially set as true. Whenever we encounter space we set the boolean variable is true. And if we encounter any character other than space, we will check the boolean variable, if it was set as true as copy that charter to the output string and set the boolean variable as false. If the boolean variable is set false, do nothing.
Algorithm:
1. Traverse string str. And initialize a variable v as true. 2. If str[i] == ' '. Set v as true. 3. If str[i] != ' '. Check if v is true or not. a) If true, copy str[i] to output string and set v as false. b) If false, do nothing.
C++
// C++ program to find the string // which contain the first character // of each word of another string. #include<bits/stdc++.h> using namespace std; // Function to find string which has // first character of each word. string firstLetterWord(string str) { string result = ""; // Traverse the string. bool v = true; for (int i = 0; i < str.length(); i++) { // If it is space, set v as true. if (str[i] == ' ') v = true; // Else check if v is true or not. // If true, copy character in output // string and set v as false. else if (str[i] != ' ' && v == true) { result.push_back(str[i]); v = false; } } return result; } // Driver code int main() { string str = "neveropen for neveropen"; cout << firstLetterWord(str); return 0; } |
gfg
Output:
gfg
Time Complexity: O(n)
Space Complexity: O(1), if space of storing resultant string is taken in account it will be O(n).
Another Approach:
In this approach, we will first split the input string based on the spaces. The spaces in the strings can be matched using a regular expression. The split strings are stored in an array of strings. Then we can simply add the first character of each split string in the result.
C++
// C++ implementation of the above // approach #include <bits/stdc++.h> using namespace std; string processWords(char *input) { /* We are splitting the input based on spaces (s)+ : this regular expression will handle scenarios where we have words separated by multiple spaces */ char *p; vector<string> s; p = strtok(input, " "); while (p != NULL) { s.push_back(p); p = strtok(NULL, " "); } string charBuffer; for (string values : s) /* charAt(0) will pick only the first character from the string and append to buffer */ charBuffer += values[0]; return charBuffer; } // Driver code int main() { char input[] = "neveropen for neveropen"; cout << processWords(input); return 0; } |
gfg
Method: Using Recursion
C++
#include<bits/stdc++.h> using namespace std; // Function to find string which has first character of each word. string firstLetterWord(string str, int index, string result) { // base case: if index is out of range, return the modified string if (index == str.length()) return result; // check for space if (str[index] == ' ') { // copy the character to result string result.push_back(str[index+1]); } // recursively call the function for the next index return firstLetterWord(str, index + 1, result); } // Driver code int main() { string str = "neveropen for neveropen"; cout<< firstLetterWord(str, 0, "g"); return 0; } //This code is contributed by Vinay Pinjala. |
gfg
Time Complexity: O(n)
Auxiliary Space: O(n)
Please refer complete article on String containing the first letter of every word in a given string with spaces for more details!
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