[Jeehay28] Week 4

merge-two-sorted-lists/Jeehay28.js

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+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} list1
+ * @param {ListNode} list2
+ * @return {ListNode}
+ */
+
+// Time Complexity: O(m + n)
+// Space Complexity: O(m + n)  
+
+var mergeTwoLists = function(list1, list2) {
+
+
+    if(!(list1 && list2)) {
+        return list1 || list2;
+    }
+
+    if(list1.val < list2.val) {
+        list1.next = mergeTwoLists(list1.next, list2);
+        return list1;
+    } else {
+        list2.next = mergeTwoLists(list1, list2.next);
+        return list2;
+    }
+
+};
+
+

missing-number/Jeehay28.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+
+// *** Guided approach 2: bitwise operations and avoids potential overflow issues with very large sums
+// XOR method
+// Time complexity: O(n)(two loops: one for numbers 0 to n  and one for array elements)
+// Space complexity: O(1)
+
+var missingNumber = function (nums) {
+  // XOR with itself results in 0 : a xor a = 0
+  // XOR with 0 results in the number itself : a xor 0 = a
+  // XOR is commutative and associative
+
+  const n = nums.length;
+
+  let xor = 0;
+
+  for (let i = 0; i <= n; i++) {
+    xor ^= i;
+  }
+
+  for (any of nums) {
+    xor ^= any;
+  }
+
+  return xor;
+};
+
+// *** Guided approach 1: simplicity and clarity
+// Gauss' Formula (Sum of First n Numbers): n*(n+1) / 2
+// Time complexity: O(n)
+// Space complexity: O(1)
+// var missingNumber = function (nums) {
+//   const n = nums.length;
+//   const expectedSum = (n * (n + 1)) / 2;
+//   const actualSum = nums.reduce((acc, cur) => acc + cur, 0); // O(n)
+
+//   const missingNum = expectedSum - actualSum;
+
+//   return missingNum;
+// };
+
+// *** My own approach
+// Time complexity: O(n^2)
+// Space complexity: O(n)
+// var missingNumber = function (nums) {
+
+//     let distinctNums = new Set([]);
+
+//     for (any of nums) {
+//         if (distinctNums.has(any)) {
+//             return
+//         } else {
+//             distinctNums.add(any)
+//         }
+//     }
+
+//     const n = distinctNums.size;
+
+//     for (let i = 0; i <= n; i++) {
+//         if (!nums.includes(i)) {
+//             return i;
+//         }
+//     }
+
+// };