Fix getLexicographicallySmallest

Author: RunDevelopment

lib/rules/sort-character-class-elements.ts

@@ -12,6 +12,7 @@ import { mention } from "../utils/mention"
 import type { ReadonlyFlags } from "regexp-ast-analysis"
 import { toUnicodeSet } from "regexp-ast-analysis"
 import type { ReadonlyWord } from "refa"
+import { getLexicographicallySmallest } from "../utils/lexicographically-smallest"
 
 type CharacterClassElementKind =
     | "\\w"
@@ -70,11 +71,7 @@ function getLexicographicallySmallestFromElement(
         node.type === "CharacterSet" && node.negate
             ? toUnicodeSet({ ...node, negate: false }, flags)
             : toUnicodeSet(node, flags)
-    const minimumWords: ReadonlyWord[] = [
-        ...(us.chars.isEmpty ? [] : [[us.chars.ranges[0].min]]),
-        ...us.accept.words,
-    ]
-    return minimumWords.sort(compareWords).shift() || []
+    return getLexicographicallySmallest(us) || []
 }
 
 /**

lib/utils/lexicographically-smallest.ts

@@ -0,0 +1,88 @@
+import type { Word } from "refa"
+import type { JS } from "refa"
+
+function findMin<T>(
+    array: readonly T[],
+    compare: (a: T, b: T) => number,
+): T | undefined {
+    if (array.length === 0) {
+        return undefined
+    }
+
+    let min = array[0]
+    for (let i = 1; i < array.length; i++) {
+        const item = array[i]
+        if (compare(item, min) < 0) {
+            min = item
+        }
+    }
+    return min
+}
+
+function compareWords(a: Word, b: Word): number {
+    const l = Math.min(a.length, b.length)
+    for (let i = 0; i < l; i++) {
+        const diff = a[i] - b[i]
+        if (diff !== 0) {
+            return diff
+        }
+    }
+    return a.length - b.length
+}
+
+/**
+ * Returns the lexicographically smallest word in the given set or `undefined` if the set is empty.
+ */
+export function getLexicographicallySmallest(
+    set: JS.UnicodeSet,
+): Word | undefined {
+    if (set.accept.isEmpty) {
+        return set.chars.isEmpty ? undefined : [set.chars.ranges[0].min]
+    }
+
+    const words = set.accept.wordSets.map(
+        (w): Word => w.map((c) => c.ranges[0].min),
+    )
+    return findMin(words, compareWords)
+}
+
+/**
+ * Returns the lexicographically smallest word in the given set or `undefined` if the set is empty.
+ */
+export function getLexicographicallySmallestInConcatenation(
+    elements: readonly JS.UnicodeSet[],
+): Word | undefined {
+    if (elements.length === 1) {
+        return getLexicographicallySmallest(elements[0])
+    }
+
+    let smallest: Word = []
+    for (let i = elements.length - 1; i >= 0; i--) {
+        const set = elements[i]
+        if (set.isEmpty) {
+            return undefined
+        } else if (set.accept.isEmpty) {
+            smallest.unshift(set.chars.ranges[0].min)
+        } else {
+            let words = set.accept.wordSets.map(
+                (w): Word => w.map((c) => c.ranges[0].min),
+            )
+            // we only have to consider the lexicographically smallest words with unique length
+            const seenLengths = new Set<number>()
+            words = words.sort(compareWords).filter((w) => {
+                if (seenLengths.has(w.length)) {
+                    return false
+                }
+                seenLengths.add(w.length)
+                return true
+            })
+
+            smallest = findMin(
+                // eslint-disable-next-line no-loop-func -- x
+                words.map((w): Word => [...w, ...smallest]),
+                compareWords,
+            )!
+        }
+    }
+    return smallest
+}