node_modules.graphql.jsutils.suggestionList.js.flow Maven / Gradle / Ivy

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Maven plugin for generating graphql clients

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/* @flow */
/**
 *  Copyright (c) 2015, Facebook, Inc.
 *  All rights reserved.
 *
 *  This source code is licensed under the BSD-style license found in the
 *  LICENSE file in the root directory of this source tree. An additional grant
 *  of patent rights can be found in the PATENTS file in the same directory.
 */

/**
 * 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: Array
): Array {
  const optionsByDistance = Object.create(null);
  const oLength = options.length;
  const inputThreshold = input.length / 2;
  for (let i = 0; i < oLength; i++) {
    const distance = lexicalDistance(input, options[i]);
    const threshold = Math.max(inputThreshold, options[i].length / 2, 1);
    if (distance <= threshold) {
      optionsByDistance[options[i]] = 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.
 *
 * 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(a, b) {
  let i;
  let j;
  const d = [];
  const aLength = a.length;
  const bLength = b.length;

  for (i = 0; i <= aLength; i++) {
    d[i] = [ i ];
  }

  for (j = 1; j <= bLength; j++) {
    d[0][j] = j;
  }

  for (i = 1; i <= aLength; i++) {
    for (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];
}