


Given a string “STR”, you need to remove spaces from the string “STR” and rewrite in the Pascal case. Your task is to return the string “STR”.
In the Pascal case writing style we don’t have spaces between words and all words begin with uppercase letters including the first word.
Note:
Input string “STR” will only consist of lowercase English Alphabets. The string will not contain white space at the beginning and end of the string.
The first line of input contains an integer ‘T’ denoting the number of test cases to run. Then the test case follows.
The first and the only line of each test case contains the string 'STR'.
Output Format :
For each test case, return a string that is written in the Pascal case style.
Output for each test case will be printed in a new line.
Note:
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 5
1 <= |STR| <= 10^5
Where |STR| denotes the length of “STR”
Time Limit: 1sec
2
coding ninja is great
i am pascal case
CodingNinjaIsGreat
IAmPascalCase
In the first case,
We have four words “coding”, “ninja, “is”, “great”. After we remove the space between these words and change the first letter of each word to an uppercase letter. the resulting string will be “CodingNinjaIsGreat” in the pascal case.
In the second case,
We have four words “i”, “am”, “pascal”, “case”. After we remove the space between these words and change the first letter of each word to an uppercase letter. the resulting string will be “IAmPascalCase” in the pascal case.
2
aba cbs llb
a b c d
AbaCbsLlb
ABCD
How can you find the first letter of each word?
We can iterate over the string “STR” and maintain the resultant string in “answer”. On each character check if that character is white space (‘ ’) then change the next character to uppercase and do not include white space in the string “answer”, else append that character in the string “answer”.
Algorithm:
O(N), where ‘N’ denotes the length of STR.
Since we iterate through the string “STR” only once, the time complexity is O(N).
O(N), where ‘N’ denotes the length of STR.
Since we store at most ‘N’ characters in the string “answer”, the space complexity is O(N).