Given a string ‘str’, two integers ‘k’ and ‘p’. The task is to count all the sub-strings of ‘str’ having exactly ‘k’ characters that have ASCII value greater than ‘p’.
Input: str = “abcd”, k=2, p=98
Only the characters ‘c’ and ‘d’ have ASCII values greater than 98,
And, the sub-strings containing them are “abcd”, “bcd” and “cd”.
Input: str = “sabrcd”, k=5, p=80
Approach: We just need to iterate over all indices and take all possible length sub-strings and then just check whether the sub-string has exactly ‘k’ characters that have ASCII value greater than ‘p’.
Typecasting a character to int will give us it’s ASCII value i.e.
int ascii = (int) char.
Below is the implementation of the above approach:
- Find number of substrings of length k whose sum of ASCII value of characters is divisible by k
- Permutation of a string with maximum number of characters greater than its adjacent characters
- Count of alphabets having ASCII value less than and greater than k
- Count the number of words having sum of ASCII values less than and greater than k
- Count number of substrings with numeric value greater than X
- Average of ASCII values of characters of a given string
- Count substrings with same first and last characters
- Count pairs of characters in a string whose ASCII value difference is K
- Convert all lowercase characters to uppercase whose ASCII value is co-prime with k
- Count characters in a string whose ASCII values are prime
- Queries for frequencies of characters in substrings
- Program to find the product of ASCII values of characters in a string
- Find the sum of the ascii values of characters which are present at prime positions
- Count distinct substrings that contain some characters at most k times
- Recursive solution to count substrings with same first and last characters
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.