From ec2d00ca9bb9aad510a2bd1640265d20779893e4 Mon Sep 17 00:00:00 2001
From: Fabian Schmidt <fabschmidt96@gmail.com>
Date: Mon, 16 Sep 2024 12:42:55 +0200
Subject: [PATCH] solved 24

---
 src/bin/problem_24.rs | 57 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 57 insertions(+)
 create mode 100644 src/bin/problem_24.rs

diff --git a/src/bin/problem_24.rs b/src/bin/problem_24.rs
new file mode 100644
index 0000000..eccaa8c
--- /dev/null
+++ b/src/bin/problem_24.rs
@@ -0,0 +1,57 @@
+use std::error::Error;
+
+fn main() {
+    let n = 1_000_000;
+    let digits = vec![1, 0, 2, 3, 4, 5, 6, 7, 8, 9];
+    let perm = permutation(digits, n).expect("Should return ok").iter().map(|&digit| digit.to_string()).collect::<Vec<String>>().join("");
+    println!("{perm}");
+    let digits = vec!["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];
+    //for n in 1..=factorial(digits.len()) {
+        let perm = permutation(digits.clone(), n).expect("Should return ok").join("");
+        println!("{perm}");
+    //}
+    let digits = vec!["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"];
+    let perm = permutation(digits.clone(), n).expect("Should return ok").join("");
+    println!("{perm}");
+}
+
+/// Explanation
+///
+/// there are 10! possible permutations
+/// for each first number there are 9!, for each first 2 numbers 8!, etc.
+/// we check how many times we have 9! permutations before we're over 1_000_000
+/// aka. 1000000 / 9!
+/// we take the remainder and check how many times we have 8! before we?re over it
+/// (1000000 % 9!) 8!
+/// etc.
+/// every iteration we remove the digit by the idx from the original permutation
+/// we only check for 999999 permutations because we already have the first one
+///
+fn permutation<T: Copy + Ord>(mut digits: Vec<T>, nth: usize) -> Result<Vec<T>, Box<dyn Error>> {
+    digits.sort();
+    if nth == 1 {
+        return Ok(digits);
+    }
+    if nth > factorial(digits.len()) || nth == 0 {
+        return Err(Box::from("Out of bounds"));
+    }
+    let mut perm = Vec::new();
+    let num_unique_digits = digits.len();
+    let mut remainder = nth - 1;
+    for idx in 1..=digits.len() {
+        let permutations = remainder / factorial(num_unique_digits-idx);
+        remainder = remainder % factorial(num_unique_digits-idx);
+        perm.push(digits[permutations]);
+        digits.remove(permutations);
+    }
+    Ok(perm)
+}
+
+// number of permutations is n! where n is the number of digits
+fn factorial(num: usize) -> usize {
+    let mut fact = 1;
+    for n in 1..=num {
+        fact = fact * n;
+    }
+    fact
+}
\ No newline at end of file