| 1 | /** |
| 2 | * Simple Levenshtein distance implementation (small N, no perf worries) |
| 3 | */ |
| 4 | const levenshtein = (a: string, b: string): number => { |
| 5 | const m = a.length |
| 6 | const n = b.length |
| 7 | const dp: Array<Array<number>> = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0)) |
| 8 | for (let i = 0; i <= m; i++) dp[i][0] = i |
| 9 | for (let j = 0; j <= n; j++) dp[0][j] = j |
| 10 | for (let i = 1; i <= m; i++) { |
| 11 | for (let j = 1; j <= n; j++) { |
| 12 | const cost = a[i - 1] === b[j - 1] ? 0 : 1 |
| 13 | dp[i][j] = Math.min( |
| 14 | dp[i - 1][j] + 1, |
| 15 | dp[i][j - 1] + 1, |
| 16 | dp[i - 1][j - 1] + cost |
| 17 | ) |
| 18 | } |
| 19 | } |
| 20 | return dp[m][n] |
| 21 | } |
| 22 | |
| 23 | /** |
| 24 | * Return suggestions with minimum distance found, up to distance 2. |
| 25 | * |
| 26 | * @internal |
| 27 | */ |
| 28 | export const suggest = (input: string, candidates: ReadonlyArray<string>): ReadonlyArray<string> => { |
| 29 | const distances = candidates |
| 30 | .map((c) => [levenshtein(input, c), c] as const) |
| 31 | .filter(([d]) => d <= 2) // Stricter threshold |
| 32 | .sort(([a], [b]) => a - b) |
| 33 | |
| 34 | if (distances.length === 0) return [] |
| 35 | |
| 36 | // Only return suggestions with the minimum distance found |
| 37 | const minDistance = distances[0][0] |
| 38 | return distances |
| 39 | .filter(([d]) => d === minDistance) |
| 40 | .map(([, c]) => c) |
| 41 | } |