✨ [mock] Added solution without logs

Author: enkr1

mock_interviews/almost_anagram/cleaned_solution.js

@@ -0,0 +1,53 @@
+const processAlmostAnagramPair = (list) => {
+  let result = new Map();
+
+  for (let i = 0; i < list.length; i++) {
+    for (let k = 0; k < list.length; k++) {
+      if (i !== k) { // Ignore the same node
+        let curr = list[i];
+        result.set(curr,
+          (result.has(curr) ? result.get(curr) : 0) + (isAlmostAnagram(curr, list[k]) ? 1 : 0))
+      }
+    }
+  }
+
+  return result;
+}
+
+const isAlmostAnagram = (item1, item2) => {
+  if (item1.length !== item2.length) return null;
+
+  let char = item1.split(""), count = 0;
+  let map1 = createMap(item1), map2 = createMap(item2);
+
+  for (let i = 0; i < char.length; i++) {
+    // Here is a trick, as long as they have the same value AND ONLY if both of them have values (appeared) more than 0.
+    // As we are doing deduction below - this is to handle cases like repeating character.
+    if ((map1.has(char[i]) && map2.has(char[i]))
+      && (0 < map1.get(char[i])
+        && 0 < map2.get(char[i]))
+    ) {
+      map1.set(char[i], map1.get(char[i]) - 1)
+      map2.set(char[i], map2.get(char[i]) - 1)
+      count++;
+    }
+  }
+
+  // Almost anagram => -1 | 1 word different
+  return (1 === (char.length - count));
+}
+
+const createMap = (item) => {
+  let map = new Map();
+  if (undefined === item) return map;
+
+  let char = item.split("");
+
+  for (let i = 0; i < char.length; i++) {
+    map.set(char[i], ((map.has(char[i])) ? map.get(char[i]) : 0) + 1)
+  }
+
+  return map;
+}
+
+processAlmostAnagramPair(["race", "beat", "brass", "grass", "sabre", "back", "chat", "cabs", "saga", "pack"]);