Advent of Code 2016 Day 3
Part 1
Three sums and three comparisons
I need to determine which three-sided shapes are valid triangles.
The proof for a valid triangle, per the instructions:
the sum of any two sides must be larger than the remaining side
This seems like the best - perhaps only? - approach:
|\
| \
| \
A| \ B
| \
|_____\
C
A + B > C?
A + C > B?
B + C > A?
My algorithm in JavaScript:
input.reduce((valids, triangle) => {
let [A,B,C] = [...triangle.matchAll(/\d+/g)].map(el => +el[0])
return valids += (
A + B > C &&
A + C > B &&
B + C > A
) ? 1 : 0
}, 0)
It generated the correct answer for Part 1!
Part 2
Columns instead of rows
A fun twist that will require nested for loops instead of a single reduce().
My algorithm as pseudocode:
Extract the digits from each line
Generate a 3-element array in place of the string
Set valid count as 0
For each 3-element array except the last two, skipping two each time
For each element in the array
Generate a 3-element array containing the numbers in the same position as the element in the array...from the current and next two arrays
Increment valid count by 1 only if each pair of side lengths is greater than the non-included side
Return valid count
My algorithm in JavaScript:
let sides = input.map(
line => [...line.matchAll(/\d+/g)].map(el => +el[0])
)
let valids = 0
for (let row = 0; row < sides.length - 2; row += 3) {
for (let col = 0; col < 3; col++) {
let [A,B,C] = sides.slice(row, row + 3).map(el => el[col])
valids += (
A + B > C &&
A + C > B &&
B + C > A
) ? 1 : 0
}
}
return valids
It generated the correct answer for Part 2!
I did it!!
- I solved both parts!
- I leveraged my growing familiarity with
regex,reduce()and several array manipulation techniques! - With the exception of Day 11, I continue a 2-star streak since Day 22!
Top comments (0)