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node_modules/graphql/jsutils/suggestionList.js.flow

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+// @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];
+}