Feb 7 - Daily Programming Diary

1:30

Yeah it's early but I thought I should complete my work on the HackOn website ASAP. Time to update the Modal, and past workshops. The most important thing is to find the data related to workshops. Since I just received the names of the workshops, I'll be going to their LinkedIn to search for content related to speaker as well as the workshop description.

3:30

Finally updated the workshop data. To get all the information was a tough job along with some UI fixes. Although I did just encountered another bug. The modal can be closed easily on the PC by clicking the backdrop but on mobile phones, the backdrop is not visible, so there is a need for a close button. I'll just reuse the add-calender icon and rotate it to look like a cross.

4:00

Took some time to position the button in the modal with some position hacks, DAMN! But it's finally done.

Time for sleep after I watch some Lucifer!

22:30

Let's start today's Leetcode challenge. The question seems to be pretty simple. The approach is to maintain a vector that stores the distance from left, and then use that to find whether the distance is minimum from left or right, and by how much.

Copyshortest-distance-to-a-character.cpp: copy code to clipboard
1vector<int> shortestToChar(string s, char c) {
2    int a = 100000;
3    vector<int> left, right(s.length(), 0);
4    for (int i = 0; i < s.length(); i++) {
5        a++;
6        if (s[i] == c) {
7            a = 0;
8        }
9        left.push_back(a);
10    }
11    a = 100000;
12    for (int i = s.length() - 1; i >= 0; i--) {
13        a++;
14        if (s[i] == c) {
15            a = 0;
16        }
17        right[i] = min(a, left[i]);
18    }
19
20    return right;
21}

The complexity is O(N) and space complexity is O(N) where N is the length of the string.

Bye!

EOD Checklist

• Leetcode February Challenge
• This blog