LC 1404. Number of Steps to Reduce a Number in Binary Representation …

…to One (Rust BM)

README.md

@@ -495,6 +495,7 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 | [1383. Maximum Performance of a Team][lc1383]                                                 | 🔴 Hard    | [![python](res/py.png)][lc1383py]                                                                     |
 | [1396. Design Underground System][lc1396]                                                     | 🟠 Medium  | [![rust](res/rs.png)][lc1396rs]                                                                       |
 | [1402. Reducing Dishes][lc1402]                                                               | 🔴 Hard    | [![python](res/py.png)][lc1402py] [![rust](res/rs.png)][lc1402rs]                                     |
+| [1404. Number of Steps to Reduce a Number in Binary Representation to One][lc1404]            | 🟠 Medium  | [![rust](res/rs.png)][lc1404rs]                                                                       |
 | [1406. Stone Game III][lc1406]                                                                | 🔴 Hard    | [![rust](res/rs.png)][lc1406rs]                                                                       |
 | [1416. Restore The Array][lc1416]                                                             | 🔴 Hard    | [![python](res/py.png)][lc1416py] [![rust](res/rs.png)][lc1416rs]                                     |
 | [1423. Maximum Points You Can Obtain from Cards][lc1423]                                      | 🟠 Medium  | [![python](res/py.png)][lc1423py]                                                                     |
@@ -1814,6 +1815,8 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 [lc1402]: https://leetcode.com/problems/reducing-dishes/
 [lc1402py]: leetcode/reducing-dishes.py
 [lc1402rs]: leetcode/reducing-dishes.rs
+[lc1404]: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one/
+[lc1404rs]: leetcode/number-of-steps-to-reduce-a-number-in-binary-representation-to-one.rs
 [lc1406]: https://leetcode.com/problems/stone-game-iii/
 [lc1406rs]: leetcode/stone-game-iii.rs
 [lc1416]: https://leetcode.com/problems/restore-the-array/

leetcode/number-of-steps-to-reduce-a-number-in-binary-representation-to-one.rs

@@ -0,0 +1,71 @@
+// 1404. Number of Steps to Reduce a Number in Binary Representation to One
+// 🟠 Medium
+//
+// https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one/
+//
+// Tags: String - Bit Manipulation
+
+struct Solution;
+impl Solution {
+    /// Iterate from the back simulating the addition operation, when we find the first one, the
+    /// carry will become 1. When carry + the current digit result in an odd value, we need to do 2
+    /// operations, otherwise we can just shift right, one operation.
+    ///
+    /// Time complexity: O(n) - Where n is the number of characters in the input.
+    /// Space complexity: O(1) - Constant extra memory used, an array with 2 i32 values.
+    ///
+    /// Runtime 0 ms Beats 100%
+    /// Memory 2.13 MB Beats 100%
+    pub fn num_steps(s: String) -> i32 {
+        s[1..]
+            .chars()
+            .map(|c| if c == '1' { 1 } else { 0 })
+            .rev()
+            .fold([0, 0], |[res, carry], d| {
+                // if (d ^ carry) & 1 == 1 {
+                if (d + carry) % 2 == 1 {
+                    [res + 2, 1]
+                } else {
+                    [res + 1, carry]
+                }
+            })
+            .into_iter()
+            .sum::<i32>()
+    }
+}
+
+// Tests.
+fn main() {
+    let tests = [
+        ("1", 0),
+        ("10", 1),
+        ("1101", 6),
+        ("11111111111111111111111111", 27),
+        ("111111111000001111111111110000011111", 47),
+    ];
+    println!("\n\x1b[92m» Running {} tests...\x1b[0m", tests.len());
+    let mut success = 0;
+    for (i, t) in tests.iter().enumerate() {
+        let res = Solution::num_steps(t.0.to_string());
+        if res == t.1 {
+            success += 1;
+            println!("\x1b[92m✔\x1b[95m Test {} passed!\x1b[0m", i);
+        } else {
+            println!(
+                "\x1b[31mx\x1b[95m Test {} failed expected: {:?} but got {}!!\x1b[0m",
+                i, t.1, res
+            );
+        }
+    }
+    println!();
+    if success == tests.len() {
+        println!("\x1b[30;42m✔ All tests passed!\x1b[0m")
+    } else if success == 0 {
+        println!("\x1b[31mx \x1b[41;37mAll tests failed!\x1b[0m")
+    } else {
+        println!(
+            "\x1b[31mx\x1b[95m {} tests failed!\x1b[0m",
+            tests.len() - success
+        )
+    }
+}