Add neetcode discourse video LC easy solution

Author: nwthomas

src/leetcode/easy/easy.md

@@ -10,6 +10,7 @@
 -   [Climbing Stairs](climbing-stairs)
 -   [Contains Duplicate](contains-duplicate)
 -   [Counting Bits](counting-bits)
+-   [Find Minimum Operations to Make All Elements Divide by Three](find-minimum-operations-to-make-all-elements-divide-by-three)
 -   [Invert Binary Tree](invert-binary-tree)
 -   [Is Palindrome](is-palindrome)
 -   [Is Same Tree](is-same-tree)

src/leetcode/easy/find-minimum-operations-to-make-all-elements-divide-by-three/findMinimumOperationsToMakeAllElementsDivideByThree.js

@@ -0,0 +1,39 @@
+/*
+https://leetcode.com/problems/find-minimum-operations-to-make-all-elements-divisible-by-three
+
+You are given an integer array nums. In one operation, you can add or subtract 1 from any element of nums.
+
+Return the minimum number of operations to make all elements of nums divisible by 3.
+
+Example 1:
+Input: nums = [1,2,3,4]
+Output: 3
+
+Explanation:
+All array elements can be made divisible by 3 using 3 operations:
+Subtract 1 from 1.
+Add 1 to 2.
+Subtract 1 from 4.
+
+Example 2:
+Input: nums = [3,6,9]
+Output: 0
+
+Constraints:
+1 <= nums.length <= 50
+1 <= nums[i] <= 50
+*/
+
+function minimumOperations(nums) {
+    let ops = 0;
+
+    nums.forEach((num) => {
+        if ([1, 2].includes(num % 3)) {
+            ops += 1;
+        }
+    });
+
+    return ops;
+}
+
+module.exports = { minimumOperations };

src/leetcode/easy/find-minimum-operations-to-make-all-elements-divide-by-three/findMinimumOperationsToMakeAllElementsDivideByThree.test.js

@@ -0,0 +1,13 @@
+const {
+    minimumOperations,
+} = require("./findMinimumOperationsToMakeAllElementsDivideByThree");
+
+describe("minimumOperations()", () => {
+    test("returns the minimum number of operations to make all elements divisible by 3", () => {
+        expect(minimumOperations([1, 2, 3, 4])).toBe(3);
+    });
+
+    test("returns the minimum number of operations to make all elements divisible by 3", () => {
+        expect(minimumOperations([3, 6, 9])).toBe(0);
+    });
+});