# Leetcode - Episode 9 - Trudging Through (3x E)

January 09, 2019 in Algorithms

The three-a-day streak nearly ended today.

I came close to completing a couple of Medium problems but I didn’t want to brute force the answers. I have a set of unwritten rules about what I’m allowed to look up in a book or google. All solutions need to be within the range of ‘optimal’ as well.

The goal is to get better at solving problems and algorithm analysis — not to get the green text that reads *Accepted*. I ended up searching for some Easys to complete so I could get to bed on time.

### 169. Majority Element

Problem: Given an array of size `n`

find the majority element that appears more than `n/2`

times.

I wrote about the algoritm that solves this problem for the first post on this blog. The *Boyer–Moore majority vote algorithm*.

Let’s take a look at its linear runtime and constant space goodness.

```
class Solution:
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
counter = 1
current = nums[0]
for i in nums[1:]:
if i != current:
counter -= 1
if counter < 0:
current = i
counter = 1
else:
counter += 1
return current
```

Its all about tracking who is in front of the pack and swapping when a challenger comes along.

Runtime complexity: `O(n)`

.

Spacetime complexity: `O(1)`

.

### 520. Detect Capital

Problem: Judge whether a word uses correct capitals.

The three types of correctness: `'IRE'`

, `'topcoder'`

, and `'Bing'`

.

I wanted to test the runtime between using Python’s built-ins and a custom solution for Leetcode’s test cases.

So, the complex way:

```
class Solution:
def detectCapitalUse(self, word):
"""
:type word: str
:rtype: bool
"""
def is_cap(c):
return ord(c) < 97
if is_cap(word[0]) and len(word) > 1 and is_cap(word[1]):
for c in word:
if not is_cap(c):
return False
elif is_cap(word[0]):
for c in word[1:]:
if is_cap(c):
return False
else:
for c in word[1:]:
if is_cap(c):
return False
return True
```

This had a runtime of 76ms.

Instead, we could have just written:

```
class Solution:
def detectCapitalUse(self, word):
"""
:type word: str
:rtype: bool
"""
return word.isupper() or word.islower() or word.istitle()
```

This *much* shorter version had a runtime of .. 76ms. The exact same result!

I’m fairly sure that as `n`

tends to infinity, the more complex solution will surely win out.

Runtime complexity: `O(n)`

.

Space complexity: `O(1)`

.

### 620. Not Boring Movies

Problem: Given a SQL table `cinema`

find movies with an odd numbered ID and a description that is not (literally) ‘boring’.

Not much to comment on this one. I read through the discussion board and some people found that using `(not .. )`

for their test expressions ran slower but I suppose that will be down to implementation details.

```
# Write your MySQL query statement below
SELECT * FROM cinema WHERE id % 2 != 0 AND description != 'boring' ORDER BY rating DESC;
```

Runtime complexity: `O(n)`

.

Spacetime complexity: `O(n)`

.

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