Hey-yo people!
Welcome to yet another edition (day 20) of my coding diary. I am sure, you might be having some thoughts about what I do here. So, every day I solve a challenge, sharing some interesting approaches + my learning.
Well, let's jump straight to the problem statement for the day!
Problem of the day - To Lower Case
Tag - Easy
Given a string s
, return the string after replacing every uppercase letter with the same lowercase letter.
Example 1:
Input: s = "Hello"
Output: "hello"
I don't even need to re-iterate the problem. It's the simple challenge one can ever get. Hold on, I know it can be solved within seconds but trust me, we are going to discuss some really cool insights.
I have a couple of solutions for this problem -
The very first, simplest solution
class Solution {
public:
string toLowerCase(string s) {
for(int i=0;i<s.length();i++) {
s[i] = tolower(s[i]);
}
return s;
}
};
I am sorry, don't punish me for using the built-in function.
ASCII comes to rescue
class Solution {
public:
string toLowerCase(string s) {
for(int i=0;i<s.length();i++) {
if(int(s[i]) >= 65 && int(s[i]) <= 90) s[i] += 32;
}
return s;
}
};
Pretty straightforward!
Tip for C++ beginners: Use int() to convert char to decimal and use char() to convert, a number to ASCII Char.
I usually, struggle to remember the exact ASCII code of A
, Z
,a
,z
. So, I did this -
class Solution {
public:
string toLowerCase(string s) {
for(int i=0;i<s.length();i++) {
if(s[i] >= 'A' && s[i] <= 'Z') s[i] += 32;
}
return s;
}
};
Ahh, Some times I don't even remember the difference is 32
. So,
if(s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
Here is the hilarious approach to optimize it further.
As, to optimize the comparison, instead of using -
if(s[i] >= 'A' && s[i] <= 'Z')
we can do,
if(s[i] <= 'Z' && s[i] >= 'A')
So, the thought is, for every lower case letter, it will get rejected in the first check itself, of the if
condition. One comparison saved! (drum rolls please)
I know I am being too picky here but that's how I optimized it further. Not to mention, my runtime beats 100% of the submissions.
One guy solved it using bit manipulation. Insane!
You might like previous editions of my coding diary
- Day #19 - N-Repeated Element in Size 2N Array.
- Day #18 - Count Negative Numbers in a Sorted Matrix.
- Day #17 - Sum of Unique Elements.
- Day #16 - Best Time to Buy and Sell Stock.
- Day #15 - Count Number of Pairs With Absolute Difference K.
- Day #14 - Minimum Number of Operations to Move All Balls to Each Box.
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