feat: 18 - maximum path sum I

readme.md

@@ -11,7 +11,7 @@ I originally started the challenge in [the-honk](https://github.com/newtykins/th
 
 ## Challenge Completion
 
-### <!-- completed -->21<!-- completed --> out of 100 public challenges completed.
+### <!-- completed -->22<!-- completed --> out of 100 public challenges completed.
 
 - [x] 1 - [Multiples of 3 or 5](src/bin/1.rs)
 - [x] 2 - [Even Fibonacci numbers](src/bin/2.rs)
@@ -30,7 +30,7 @@ I originally started the challenge in [the-honk](https://github.com/newtykins/th
 - [x] 15 - [Lattice paths](src/bin/15.rs)
 - [x] 16 - [Power digit sum](src/bin/16.rs)
 - [x] 17 - [Number letter counts](src/bin/17.rs)
-- [ ] 18 - Maximum path sum I
+- [x] 18 - [Maximum path sum I](src/bin/18.rs)
 - [ ] 19 - Counting Sundays
 - [ ] 20 - Factorial digit sum
 - [ ] 21 - Amicable numbers

src/bin/17.rs

@@ -6,9 +6,8 @@ If all the numbers from 1 to 1000 (one thousand) inclusive were written out in w
 NOTE:  Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
 */
 
-use std::ops::Div;
-
 use phf::phf_map;
+use std::ops::Div;
 
 const TRANSLATIONS: phf::Map<u16, &'static str> = phf_map! {
     0u16 => "",
@@ -58,7 +57,10 @@ fn number_to_words(number: usize) -> String {
     n -= hundreds * 100;
 
     if hundreds > 0 {
-        out = format!("{out}{}hundred", TRANSLATIONS.get(&(hundreds as u16)).unwrap());
+        out = format!(
+            "{out}{}hundred",
+            TRANSLATIONS.get(&(hundreds as u16)).unwrap()
+        );
     }
 
     // Tens
@@ -67,11 +69,11 @@ fn number_to_words(number: usize) -> String {
 
     if tens > 0 {
         let translation: &str;
-        let and = if hundreds > 0 {"and"} else {""};
+        let and = if hundreds > 0 { "and" } else { "" };
 
         if n % 10 > 0 && tens == 1 {
             translation = TRANSLATIONS.get(&((10 + (n % 10)) as u16)).unwrap();
-            
+
             n -= n % 10;
         } else {
             translation = TRANSLATIONS.get(&(tens as u16 * 10)).unwrap();
@@ -82,7 +84,7 @@ fn number_to_words(number: usize) -> String {
 
     // Remainder
     if n > 0 {
-        let and = if hundreds > 0 && tens <= 0 {"and"} else {""};
+        let and = if hundreds > 0 && tens <= 0 { "and" } else { "" };
         out = format!("{out}{and}{}", TRANSLATIONS.get(&(n as u16)).unwrap());
     }
 
@@ -104,5 +106,8 @@ fn letter_count(upper_bound: usize) -> usize {
 fn main() {
     let sum = letter_count(1000);
 
-    println!("{} letters would be used to write out all of the numbers from 1 to 1000!", sum);
+    println!(
+        "{} letters would be used to write out all of the numbers from 1 to 1000!",
+        sum
+    );
 }

src/bin/18.rs

@@ -0,0 +1,62 @@
+/*
+Problem 18 - Maximum path sum I
+
+By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
+3     7   4
+2   4   6
+8 5   9   3
+That is, 3 + 7 + 4 + 9 = 23.
+Find the maximum total from top to bottom of the triangle below:
+75
+95 64
+17 47 82
+18 35 87 10
+20 04 82 47 65
+19 01 23 75 03 34
+88 02 77 73 07 63 67
+99 65 04 28 06 16 70 92
+41 41 26 56 83 40 80 70 33
+41 48 72 33 47 32 37 16 94 29
+53 71 44 65 25 43 91 52 97 51 14
+70 11 33 28 77 73 17 78 39 68 17 57
+91 71 52 38 17 14 91 43 58 50 27 29 48
+63 66 04 68 89 53 67 30 73 16 69 87 40 31
+04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
+NOTE:  As there are only 16384 routes, it is possible to solve this problem by trying every route. However,  Problem 67 , is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
+*/
+
+fn maximum_path_sum(triangle: Vec<Vec<usize>>) -> usize {
+    let mut tri = triangle.clone();
+
+    for row in (0..(triangle.len() - 1)).rev() {
+        for col in 0..(row + 1) {
+            tri[row][col] = tri[row][col] + std::cmp::max(tri[row + 1][col], tri[row + 1][col + 1]);
+        }
+    }
+
+    return tri[0][0];
+}
+
+fn main() {
+    let triangle: Vec<Vec<usize>> = vec![
+        vec![75],
+        vec![95, 64],
+        vec![17, 47, 82],
+        vec![18, 35, 87, 10],
+        vec![20, 4, 82, 47, 65],
+        vec![19, 1, 23, 75, 3, 34],
+        vec![88, 2, 77, 73, 7, 63, 67],
+        vec![99, 65, 4, 28, 6, 16, 70, 92],
+        vec![41, 41, 26, 56, 83, 40, 80, 70, 33],
+        vec![41, 48, 72, 33, 47, 32, 37, 16, 94, 29],
+        vec![53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14],
+        vec![70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57],
+        vec![91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48],
+        vec![63, 66, 4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
+        vec![4, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 4, 23],
+    ];
+    
+    let sum = maximum_path_sum(triangle);
+
+    println!("The maximum path sum of the triangle is {}!", sum);
+}