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/**
* Simple Levenshtein distance implementation (small N, no perf worries)
*/
const levenshtein = (a, b) => {
const m = a.length;
const n = b.length;
const dp = 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, candidates) => {
const distances = candidates.map(c => [levenshtein(input, c), c]).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);
};
//# sourceMappingURL=auto-suggest.js.map

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