add 343 554 scripts.

Author: everthis

343-integer-break.js

@@ -0,0 +1,34 @@
+/**
+ * @param {number} n
+ * @return {number}
+ */
+const integerBreak = function(n) {
+  if (n <= 2) return 1;
+
+  const maxArr = [];
+  for (let i = 0; i < n + 1; i++) {
+    maxArr[i] = 0;
+  }
+
+  /** For a number i: write i as a sum of integers, then take the product of those integers.
+   maxArr[i] = maximum of all the possible products */
+
+  maxArr[1] = 0;
+  maxArr[2] = 1; // 2=1+1 so maxArr[2] = 1*1
+
+  for (let i = 3; i <= n; i++) {
+    for (let j = 1; j < i; j++) {
+      /** Try to write i as: i = j + S where S=i-j corresponds to either one number or a sum of two or more numbers
+        
+        Assuming that j+S corresponds to the optimal solution for maxArr[i], we have two cases:
+        (1) i is the sum of two numbers, i.e. S=i-j is one number, and so maxArr[i]=j*(i-j)
+        (2) i is the sum of at least three numbers, i.e. S=i-j is a sum of at least 2 numbers,
+        and so the product of the numbers in this sum for S is maxArr[i-j]
+        (=maximum product after breaking up i-j into a sum of at least two integers):
+        maxArr[i] = j*maxArr[i-j]
+        */
+      maxArr[i] = Math.max(maxArr[i], j * (i - j), j * maxArr[i - j]);
+    }
+  }
+  return maxArr[n];
+};

554-brick-wall.js

@@ -0,0 +1,25 @@
+/**
+ * @param {number[][]} wall
+ * @return {number}
+ */
+const leastBricks = function(wall) {
+  const hash = {};
+  let row;
+  let rowSum = 0;
+  for (let i = 0; i < wall.length; i++) {
+    rowSum = 0;
+    row = wall[i];
+    for (let j = 0; j < row.length - 1; j++) {
+      rowSum += row[j];
+      hash[rowSum] = hash.hasOwnProperty(rowSum) ? hash[rowSum] + 1 : 1;
+    }
+  }
+  return (
+    wall.length -
+    (Object.keys(hash).length > 0
+      ? Math.max(...Object.keys(hash).map(key => hash[key]))
+      : 0)
+  );
+};
+
+console.log(leastBricks([[1], [1], [1]]));