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Leetcode Problem #709 (Easy): To Lower Case

Description:

(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

Given a string s, return the string after replacing every uppercase letter with the same lowercase letter.

Examples:

Example 1:
Input:	s = "Hello"
Output:	"hello"

Example 2:
Input:	s = "here"
Output:	"here"

Example 3:
Input:	s = "LOVELY"
Output:	"lovely"

Constraints:

• 1 <= s.length <= 100
• s consists of printable ASCII characters.

Idea:

(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

The uppercase letters from 'A' to 'Z' have ASCCII codes between 65 and 90. We can iterate through our input string (s) and as we build up our answer string (ans), if the value (n) of any character lies in that range, we can replace it with the character of value n + 32, which corresponds to the lowercase version of the uppercase letter.

• Time Complexity: O(N) where N is the length of s
• Space Complexity: O(1) excluding the space of the output

Javascript Code:

(Jump to: Problem Description || Solution Idea)

var toLowerCase = function(s) {
    let ans = ""
    for (let c of s) {
        let n = c.charCodeAt()
        ans += n > 64 && n < 91 ? String.fromCharCode(n + 32) : c 
    }
    return ans
};

Python Code:

(Jump to: Problem Description || Solution Idea)

class Solution:
    def toLowerCase(self, s: str) -> str:
        ans = ""
        for c in s:
            n = ord(c)
            ans += chr(n+32) if n > 64 and n < 91 else c
        return ans

Java Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
    public String toLowerCase(String s) {
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            ans.append(c > 64 && c < 91 ? (char)(c + 32) : c);
        }
        return new String(ans);
    }
}

C++ Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    string toLowerCase(string s) {
        string ans = "";
        for (auto& c : s)
            ans += c > 64 && c < 91 ? c + 32 : c;
        return ans;
    }
};