16(July) Remaining String

16. Remaining String

The problem can be found at the following link: Question Link

Problem Description

Given a string s without spaces, a character ch, and an integer count, return the substring that remains after the character ch has appeared count number of times. Assume upper case and lower case alphabets are different. Return an empty string if it is not possible or the remaining substring is empty.

Example:

Input:

s = "Thisisdemostring", ch = 'i', count = 3

Output:

ng

Explanation: The remaining substring of s after the 3rd occurrence of 'i' is "ng", hence the output is "ng".

My Approach

1. Initialization:

• Initialize a variable occ to count the occurrences of the character ch in the string s.

1. Iteration through the string:

• Iterate through each character in the string s.

• For each character, check if it matches ch.

• If it matches, increment the occ counter.

• When occ equals count, check if the next character exists in the string.

  • If it exists, return the substring starting from the next character.

  • If it does not exist, return an empty string.

1. Return:

• If the loop completes and occ is less than count, return an empty string.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(|s|), as we iterate through the string s once.

• Expected Auxiliary Space Complexity: O(1), as we use a constant amount of additional space.

Code (C++)

class Solution {
public:
    string printString(const string &str, char ch, int count) {
        int occ = 0;

        for (int i = 0; i < str.size(); i++) {
            if (str[i] == ch) {
                occ++;
            }
            if (occ == count) {
                if (i + 1 < str.size()) {
                    return str.substr(i + 1);
                } else {
                    return "";
                }
            }
        }
        return "";
    }
};

Code (Java)

class Solution {
    public String printString(String s, char ch, int count) {
        int occ = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == ch) {
                occ++;
            }
            if (occ == count) {
                if (i + 1 < s.length()) {
                    return s.substring(i + 1);
                } else {
                    return "";
                }
            }
        }
        return "";
    }
}

Code (Python)

class Solution:
    def printString(self, s, ch, count):
        occ = 0

        for i in range(len(s)):
            if s[i] == ch:
                occ += 1
            if occ == count:
                if i + 1 < len(s):
                    return s[i + 1:]
                else:
                    return ""
        return ""

Contribution and Support

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

⭐ If you find this helpful, please give this repository a star! ⭐

---

📍Visitor Count