feat: lodash chunk solution

problem/implement-lodash-chunk_en.md

@@ -1,4 +1,64 @@
+> This solution is submitted by [@CaptainOfFlyingDutchman](https://github.com/CaptainOfFlyingDutchman)
 
-There is no solution yet.
+To implement this solution we need to figure out 2 things
 
-Would you like to [contribute to the solution](https://github.com/BFEdev/BFE.dev-solutions/blob/main/problem/implement-lodash-chunk_en.md)? [Contribute guideline](https://github.com/BFEdev/BFE.dev-solutions#how-to-contribute)
+1. How many times we need to iterate through the array.
+2. Should not modify the original array, which is a good practice in itself.
+
+---
+
+We just cannot iterate through the array elements one by one, if we do that then we need to maintain additional state variables to track **how many elements** we need to iterate for before collecting them in a new array and then start again collecting then in a new iteration.
+
+Did you notice **how many elements** in last paragraph? We have this information already through the second parameter `size`. We can easily use that to iterate through the array using the convention old school `for` statement.
+
+The following code will increase the variable `i` by `size` instead of traditional `i++` syntax (that can be written like `i += 1` by the way).
+
+```javascript
+for (let i = 0; i < items.length; i += size) {
+    // do stuff
+}
+```
+
+As we should not modify the original array we can use the `Array.prototype.slice` method, that slice the array using two given indexes and returns a new array leaving original array intact (a bit of **immutability** genius right there).
+
+One thing to note that the second parameter to `slice` is optional and can be greater than the array's length, in that case `slice` will return all the remaining elements starting `start` parameter to end of the array.
+
+```javascript
+const sliced = items.slice(0, 3);
+// items before slice => [1, 2, 3, 4, 5]
+// slices => [1, 2, 3]
+// items after slice => [1, 2, 3, 4, 5]
+```
+
+The last thing to do is to capture this `slice` result in an array because that is what `chunk` needs to return, an *Array*.
+
+```javascript
+const chunks = [];
+
+chunks.push(items.slice(0, 3));
+```
+
+We also need to take care of an edge case where if user supplies `0` as the size then return an empty array `[]`, or else our loop will become an `infinite loop` because what would we get adding `zero` in the already `zero` initialized variable `i`, `0`.
+
+Here is the full implementation
+
+```javascript
+/**
+ * @param {any[]} items
+ * @param {number} size
+ * @returns {any[][]}
+ */
+function chunk(items, size) {
+  if (size < 1) {
+    return [];
+  }
+
+  const chunks = [];
+
+  for (let i = 0; i < items.length; i += size) {
+    chunks.push(items.slice(i, size + i));
+  }
+
+  return chunks;
+}
+```