Dec 3, 2018

Author: dalindev

709__easy__to-lower-case.js

@@ -0,0 +1,44 @@
+/**
+
+    Implement function ToLowerCase() that has a string parameter str, and returns the same string in lowercase.
+
+    Example 1:
+
+    Input: "Hello"
+    Output: "hello"
+    Example 2:
+
+    Input: "here"
+    Output: "here"
+    Example 3:
+
+    Input: "LOVELY"
+    Output: "lovely"
+
+ */
+
+/**
+ *
+ * O(n) time
+ * O(n) space
+ * Runtime: 48 ms, faster than 100.00%
+ */
+/**
+ * @param {string} str
+ * @return {string}
+ */
+var toLowerCase = function(str) {
+    if (!str) return str;
+    // Char Code A-Z => 65 - 90, a-z 97 - 122
+    // A -> a need + 32 to char code of A
+    let res = "";
+    for (let i = 0; i < str.length; i++) {
+        let cCode = str[i].charCodeAt(0);
+        if (65 <= cCode && cCode <= 90) {
+            res += String.fromCharCode(cCode + 32);
+        } else {
+            res += str[i];
+        }
+    }
+    return res;
+};

760__easy__find-anagram-mappings.js

@@ -0,0 +1,48 @@
+/**
+
+    Given two lists Aand B, and B is an anagram of A. B is an anagram of A means B is made by randomizing the order of the elements in A.
+
+    We want to find an index mapping P, from A to B. A mapping P[i] = j means the ith element in A appears in B at index j.
+
+    These lists A and B may contain duplicates. If there are multiple answers, output any of them.
+
+    For example, given
+
+    A = [12, 28, 46, 32, 50]
+    B = [50, 12, 32, 46, 28]
+    We should return
+    [1, 4, 3, 2, 0]
+    as P[0] = 1 because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element of A appears at B[4], and so on.
+    Note:
+
+    A, B have equal lengths in range [1, 100].
+    A[i], B[i] are integers in range [0, 10^5]
+
+ */
+/**
+ *
+ * O(n) time
+ * O(n) space
+ *
+ * Runtime: 52 ms, faster than 100.00%
+ */
+/**
+ * @param {number[]} A
+ * @param {number[]} B
+ * @return {number[]}
+ */
+var anagramMappings = function(A, B) {
+    if (!A || !B) return null;
+
+    let map = {}; // {} is faster than Map()
+    for (let i = 0; i < B.length; i++) {
+        if (!map[B[i]]) map[B[i]] = i;
+    }
+
+    let res = new Array(A.length);
+    for (let i = 0; i < A.length; i++) {
+        res[i] = map[A[i]];
+    }
+
+    return res;
+};