Solve longest-common-prefix.

source/lib.rs

@@ -1,3 +1,4 @@
+pub mod longest_common_prefix;
 pub mod palindrome_number;
 pub mod reverse_integer;
 pub mod two_sum;

source/longest_common_prefix/mod.rs

@@ -0,0 +1,71 @@
+pub fn longest_common_prefix(strings: Vec<String>) -> String {
+  // Sort the strings ascending by length.
+  let mut strings = strings.clone();
+  strings.sort_by(|a, b| a.len().cmp(&b.len()));
+
+  // Then grab the shortest string.
+  let shortest_string = match strings.first() {
+    Some(string) => string,
+    None => return String::new(),
+  };
+
+  // The shortest string must be the maximum possible prefix, so start with it.
+  let mut prefix = shortest_string.clone();
+
+  // Then loop over the shortest string's length and check if every string in
+  // the list starts with the prefix, and if they don't, pop the last character
+  // from the prefix.
+  for _ in 0..shortest_string.len() {
+    // If we find a prefix that all strings start with, we've found the longest
+    // common one.
+    if strings.iter().all(|string| string.starts_with(&prefix)) {
+      break;
+    } else {
+      prefix.pop();
+    }
+  }
+
+  // If by the end there is no longest common prefix, the prefix will be an
+  // empty string.
+  prefix
+}
+
+#[test]
+fn test_longest_common_prefix() {
+  assert_eq!(
+    longest_common_prefix(vec![
+      "doggo".to_string(),
+      "doggie".to_string(),
+      "dog".to_string(),
+    ]),
+    "dog"
+  );
+
+  assert_eq!(
+    longest_common_prefix(vec![
+      "flower".to_string(),
+      "flow".to_string(),
+      "flight".to_string(),
+    ]),
+    "fl"
+  );
+
+  assert_eq!(
+    longest_common_prefix(vec![
+      "dog".to_string(),
+      "racecar".to_string(),
+      "car".to_string(),
+    ]),
+    ""
+  );
+
+  assert_eq!(
+    longest_common_prefix(vec![
+      "dog".to_string(),
+      "racecar".to_string(),
+      "car".to_string(),
+      "".to_string()
+    ]),
+    ""
+  );
+}