| // @flow strict |
| |
| /** |
| * Given an invalid input string and a list of valid options, returns a filtered |
| * list of valid options sorted based on their similarity with the input. |
| */ |
| export default function suggestionList( |
| input: string, |
| options: $ReadOnlyArray<string>, |
| ): Array<string> { |
| const optionsByDistance = Object.create(null); |
| const inputThreshold = input.length / 2; |
| for (const option of options) { |
| const distance = lexicalDistance(input, option); |
| const threshold = Math.max(inputThreshold, option.length / 2, 1); |
| if (distance <= threshold) { |
| optionsByDistance[option] = distance; |
| } |
| } |
| return Object.keys(optionsByDistance).sort((a, b) => { |
| const distanceDiff = optionsByDistance[a] - optionsByDistance[b]; |
| return distanceDiff !== 0 ? distanceDiff : a.localeCompare(b); |
| }); |
| } |
| |
| /** |
| * Computes the lexical distance between strings A and B. |
| * |
| * The "distance" between two strings is given by counting the minimum number |
| * of edits needed to transform string A into string B. An edit can be an |
| * insertion, deletion, or substitution of a single character, or a swap of two |
| * adjacent characters. |
| * |
| * Includes a custom alteration from Damerau-Levenshtein to treat case changes |
| * as a single edit which helps identify mis-cased values with an edit distance |
| * of 1. |
| * |
| * This distance can be useful for detecting typos in input or sorting |
| * |
| * @param {string} a |
| * @param {string} b |
| * @return {int} distance in number of edits |
| */ |
| function lexicalDistance(aStr, bStr) { |
| if (aStr === bStr) { |
| return 0; |
| } |
| |
| const d = []; |
| const a = aStr.toLowerCase(); |
| const b = bStr.toLowerCase(); |
| const aLength = a.length; |
| const bLength = b.length; |
| |
| // Any case change counts as a single edit |
| if (a === b) { |
| return 1; |
| } |
| |
| for (let i = 0; i <= aLength; i++) { |
| d[i] = [i]; |
| } |
| |
| for (let j = 1; j <= bLength; j++) { |
| d[0][j] = j; |
| } |
| |
| for (let i = 1; i <= aLength; i++) { |
| for (let j = 1; j <= bLength; j++) { |
| const cost = a[i - 1] === b[j - 1] ? 0 : 1; |
| |
| d[i][j] = Math.min( |
| d[i - 1][j] + 1, |
| d[i][j - 1] + 1, |
| d[i - 1][j - 1] + cost, |
| ); |
| |
| if (i > 1 && j > 1 && a[i - 1] === b[j - 2] && a[i - 2] === b[j - 1]) { |
| d[i][j] = Math.min(d[i][j], d[i - 2][j - 2] + cost); |
| } |
| } |
| } |
| |
| return d[aLength][bLength]; |
| } |