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/**
* Simple Levenshtein distance implementation (small N, no perf worries)
*/
const levenshtein = (a: string, b: string): number => {
const m = a.length
const n = b.length
const dp: Array<Array<number>> = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0))
for (let i = 0; i <= m; i++) dp[i][0] = i
for (let j = 0; j <= n; j++) dp[0][j] = j
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
const cost = a[i - 1] === b[j - 1] ? 0 : 1
dp[i][j] = Math.min(
dp[i - 1][j] + 1,
dp[i][j - 1] + 1,
dp[i - 1][j - 1] + cost
)
}
}
return dp[m][n]
}
/**
* Return suggestions with minimum distance found, up to distance 2.
*
* @internal
*/
export const suggest = (input: string, candidates: ReadonlyArray<string>): ReadonlyArray<string> => {
const distances = candidates
.map((c) => [levenshtein(input, c), c] as const)
.filter(([d]) => d <= 2) // Stricter threshold
.sort(([a], [b]) => a - b)
if (distances.length === 0) return []
// Only return suggestions with the minimum distance found
const minDistance = distances[0][0]
return distances
.filter(([d]) => d === minDistance)
.map(([, c]) => c)
}

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