# Daily Coding Challenge #3

wkw Updated on ・2 min read

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

``````/*
Leetcode - Valid Palindrome

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true
Example 2:

Input: "race a car"
Output: false

*/

class Solution {
public:
bool isPalindrome(string s) {
int sz=(int)s.size();
if(!sz) return true;
int i=0,k=sz-1;
while(i<k){
// move to the first alphanumeric character
if(!isalnum(s[i])){
i++;
continue;
}
// move to the first alphanumeric character
if(!isalnum(s[k])){
k--;
continue;
}
// compare both characters, return false if they are not the same
if(tolower(s[i++])!=tolower(s[k--])) {
return false;
}
}
return true;
}
};

// A man, a plan, a canal: Panama
// amanaplanacanalpanama
``````
``````/*
Leetcode - First Bad Version

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example:

Given n = 5, and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version.
*/

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
int firstBadVersion(int n) {
int l=0, h=n, mid;
while(l<h){
mid=l+(h-l)/2;
// Scenario #1: isBadVersion(mid) => true

//  1 2 3 4 5 6 7 8 9
//  G G G B B B B B B       G = Good, B = Bad
//  |       |       |
// left    mid    right
if(isBadVersion(mid)) h = mid;
// Scenario #2: isBadVersion(mid) => false

// 1 2 3 4 5 6 7 8 9
// G G G G G G B B B       G = Good, B = Bad
// |       |       |
// left    mid    right
else l = mid+1;
}
return h;
}
};

static const auto io_sync_off = []() {std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();
``````

The source code is available in corresponding repo below. Star and watch for timely updates!