Back_shop/node_modules/graphql/jsutils/suggestionList.js.flow

// @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) => optionsByDistance[a] - optionsByDistance[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];
}