aoc_2 fast version

src/aoc/aoc_2.rs

@@ -1,28 +1,238 @@
-use std::io::BufRead;
+use strength_reduce::StrengthReducedU64;
-
+pub fn solve_p1(input: &str) -> u64 {
-pub fn solve_p1(input: &str ) -> u64 {
     let mut answer = 0;
     for line in input.split(',') {
-        let (num1, num2) = line.split_once('-').map(|n| {(n.0.parse::<u64>().unwrap(), n.1.parse::<u64>().unwrap())}).unwrap();
+        let (num1, num2) = line
+            .split_once('-')
+            .map(|n| (n.0.parse::<u64>().unwrap(), n.1.parse::<u64>().unwrap()))
+            .unwrap();
         for i in num1..=num2 {
             let text = i.to_string();
-            if text.len()%2 == 0 {
+            if text.len() % 2 == 0 {
-                let splits = text.split_at(text.len()/2);
+                let splits = text.split_at(text.len() / 2);
-                if splits.0==splits.1 {
+                if splits.0 == splits.1 {
-                    answer +=i;
+                    answer += i;
                 }
             }
         }
     }
 
-
     answer
 }
 
-pub fn solve_p2(input: &str ) -> u64 {
+const POW10: [u64; 10] = [
+    1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000,
+];
+
+pub fn solve_p1_f(input: &str) -> u64 {
+    let mut total_answer = 0;
+
+    for line in input.split(',') {
+        let Some((s1, s2)) = line.split_once('-') else {
+            continue;
+        };
+        let n1: u64 = s1.parse().unwrap();
+        let n2: u64 = s2.parse().unwrap();
+
+        // Iterate through possible "half lengths".
+        // u64 max is ~1.8e19, so max length is 19.
+        // We only care about even lengths: 2, 4, 6... 18.
+        // So 'half_len' goes from 1 to 9.
+        let half_len = n1.ilog10() / 2 + 1;
+        let power_of_10 = POW10[half_len as usize];
+        let multiplier = power_of_10 + 1;
+
+        // X must be a number with exactly 'half_len' digits.
+        // Range for X: [10^(d-1), 10^d - 1]
+        let x_min_digits = POW10[(half_len - 1) as usize];
+        let x_max_digits = power_of_10 - 1;
+
+        // X must also satisfy: n1 <= X * multiplier <= n2
+        // Therefore: n1/multiplier <= X <= n2/multiplier
+        // We use ceiling for lower bound and floor for upper bound.
+        let x_min_val = (n1 + multiplier - 1) / multiplier;
+        let x_max_val = n2 / multiplier;
+
+        // Find the intersection of valid digits and valid values
+        let start = x_min_digits.max(x_min_val);
+        let end = x_max_digits.min(x_max_val);
+
+        if start <= end {
+            // Sum of numbers N = X * multiplier
+            // Sum = multiplier * Sum(X for X in start..=end)
+            // Sum(start..=end) = (start + end) * count / 2
+            let count = end - start + 1;
+            let sum_x = (start + end) * count / 2;
+            total_answer += sum_x * multiplier;
+        }
+    }
+
+    total_answer
+}
+
+pub fn solve_p1_bytes(bytes: &[u8]) -> u64 {
+    let mut total_answer = 0;
+    let mut i = 0;
+    let len = bytes.len();
+
+    while i < len {
+        // 1. Parse N1 and count its digits (c1)
+        let mut n1: u64 = 0;
+        let mut c1: usize = 0;
+        while i < len {
+            let b = bytes[i];
+            if b == b'-' {
+                i += 1;
+                break;
+            }
+            n1 = n1 * 10 + (b & 0x0F) as u64;
+            c1 += 1;
+            i += 1;
+        }
+
+        // 2. Parse N2 and count its digits (c2)
+        let mut n2: u64 = 0;
+        while i < len {
+            let b = bytes[i];
+            if b == b',' {
+                i += 1;
+                break;
+            }
+            n2 = n2 * 10 + (b & 0x0F) as u64;
+            i += 1;
+        }
+
+        let half_len = (c1 + 1) >> 1; // Bitwise division by 2
+
+        // --- INLINED MATH (No function call overhead) ---
+        let power_of_10 = POW10[half_len];
+        let multiplier = power_of_10 + 1;
+
+        let x_min_digits = POW10[half_len - 1];
+        let x_max_digits = power_of_10 - 1;
+
+        // Fast ceiling division: (n1 + m - 1) / m
+        let x_min_val = (n1 + multiplier - 1) / multiplier;
+        let x_max_val = n2 / multiplier;
+
+        // Branchless max/min
+        let start = if x_min_digits > x_min_val {
+            x_min_digits
+        } else {
+            x_min_val
+        };
+        let end = if x_max_digits < x_max_val {
+            x_max_digits
+        } else {
+            x_max_val
+        };
+
+        if start <= end {
+            let count = end - start + 1;
+            // Formula: Sum = multiplier * (start + end) * count / 2
+            total_answer += multiplier * (start + end) * count / 2;
+        }
+    }
+
+    total_answer
+}
+
+pub fn solve_p1_unsafe(input: &str) -> u64 {
+    let mut total_answer = 0;
+
+    // Get raw pointers
+    let mut ptr = input.as_ptr();
+    // Calculate the end address
+    let end = unsafe { ptr.add(input.len()) };
+
+    while ptr < end {
+        // 1. Parse N1 (Unsafe, no bounds check)
+        let mut n1: u64 = 0;
+        let start_ptr = ptr;
+        loop {
+            // Read byte directly from memory
+            let b = unsafe { *ptr };
+            if b == b'-' {
+                ptr = unsafe { ptr.add(1) }; // Skip '-'
+                break;
+            }
+            // Fast ASCII conversion
+            n1 = n1 * 10 + (b & 0x0F) as u64; // b & 0x0F is faster than b - b'0'
+            ptr = unsafe { ptr.add(1) };
+        }
+
+        // Calculate digits of N1 using pointer difference (Instant)
+        // (Current ptr is at dash + 1) - (Start ptr) - 1 (for the dash)
+        let c1 = (ptr as usize - start_ptr as usize - 1) as u8;
+
+        // 2. Parse N2
+        let mut n2: u64 = 0;
+        // let start_ptr2 = ptr; // Not strictly needed unless you want c2
+        loop {
+            // Check if we hit end of string (EOF protection)
+            if ptr == end {
+                break;
+            }
+
+            let b = unsafe { *ptr };
+            if b == b',' {
+                ptr = unsafe { ptr.add(1) }; // Skip ','
+                break;
+            }
+
+            // Only valid digits here
+            n2 = n2 * 10 + (b & 0x0F) as u64;
+            ptr = unsafe { ptr.add(1) };
+        }
+
+        // --- LOGIC ---
+        // c1 is enough because we know the length difference is <= 1.
+        // If c1 is even, we check c1. If c1 is odd, we check c1 + 1 (which is c2).
+
+        // Trick: We need to check 'c1' if it's even, or 'c1+1' if c1 is odd.
+        // Let's just calculate the specific half_len we care about.
+
+        // If c1 is even (e.g. 2), we want half_len=1.
+        // If c1 is odd (e.g. 3), we want half_len=2 (because target must be len 4).
+        // Formula: half_len = (c1 + 1) / 2
+        let half_len = ((c1 + 1) >> 1) as usize;
+
+        let power_of_10 = unsafe { *POW10.get_unchecked(half_len) };
+        let multiplier = power_of_10 + 1;
+
+        let x_min_digits = unsafe { *POW10.get_unchecked(half_len - 1) };
+        let x_max_digits = power_of_10 - 1;
+
+        let x_min_val = (n1 + multiplier - 1).div_euclid(multiplier);
+        let x_max_val = n2 / multiplier;
+
+        let start = if x_min_digits > x_min_val {
+            x_min_digits
+        } else {
+            x_min_val
+        };
+        let end = if x_max_digits < x_max_val {
+            x_max_digits
+        } else {
+            x_max_val
+        };
+
+        if start <= end {
+            let count = end - start + 1;
+            total_answer += multiplier * (start + end) * count / 2;
+        }
+    }
+
+    total_answer
+}
+
+pub fn solve_p2(input: &str) -> u64 {
     let mut answer = 0;
     for line in input.split(',') {
-        let num_range = line.split('-').map(|num| { num.parse::<u64>().unwrap_or(0) }).collect::<Vec<u64>>();
+        let num_range = line
+            .split('-')
+            .map(|num| num.parse::<u64>().unwrap_or(0))
+            .collect::<Vec<u64>>();
         for i in num_range[0]..=num_range[1] {
             let text = i.to_string();
             for j in 1..=text.len() / 2 {