feat: Solve ice cream parlor

  - Hackerrank

Author: brianr482

ice-cream-parlor.js

@@ -0,0 +1,77 @@
+/**
+ * Hackerrank
+ * Sunny and Johnny like to pool their money and go to the ice cream parlor.
+ * Johnny never buys the same flavor that Sunny does. The only other rule they
+ * have is that they spend all of their money.
+ * Given a list of prices for the flavors of ice cream, select the two that
+ * will cost all of the money they have.
+ * 
+ * Constraints:
+ * 1. 2 <= m <= 10^4 (The integer m, the amount of money they have pooled)
+ * 2. 2 <= n <= 10^4 (The integer n, the number of flavors offered at the time)
+ * 3. 1 <= cost[i] <= 10^4, i ∈ [1, n]
+ * 
+ * This solution got 30 points
+ * Problem link: http://hr.gs/eez
+ */
+
+ // Complete the icecreamParlor function below.
+ /**
+  * 
+  * @param {number} m The total amount of money pooled
+  * @param {number[]} arr The list of ice creams with their price
+  */
+function icecreamParlor(m, arr) {
+  const icecreamIdx = new Map();
+  // Store the original indices
+  arr.forEach((n, i) => {
+    if (icecreamIdx.has(n)) {
+      icecreamIdx.get(n).push(i + 1);
+    } else {
+      icecreamIdx.set(n, [i + 1]);
+    }
+  });
+  // Sort array in order to use binary search
+  arr.sort((a, b) => a - b);
+  let i = 0;
+  let price = arr[0];
+  while (i < arr.length && price < m) {
+    const slicedArray = arr.slice(i);
+    // Use binary search to look for a value that meet the criteria
+    const complementIndex = binarySearch(
+      m - price, 0, slicedArray.length - 1, slicedArray
+    );
+    if (complementIndex != null) {
+      // Sort and return the response
+      return [
+        icecreamIdx.get(price).pop(),
+        icecreamIdx.get(slicedArray[complementIndex]).pop()
+      ].sort((a, b) => a - b);
+    }
+    i++;
+    price = arr[i];
+  }
+  return [];
+}
+
+/**
+ * Look for an element in array by binary search
+ * @param {numner} val The value to look for
+ * @param {number} l Left limit
+ * @param {number} r Right limit
+ * @param {number[]} arr The array where the search will be executed 
+ */
+function binarySearch(val, l, r, arr) {
+  if (l > r) {
+    return null;
+  }
+  const mid = l + Math.floor((r - l) / 2);
+  if (arr[mid] === val) {
+    return mid;
+  }
+  if(arr[mid] > val) {
+    return binarySearch(val, l, mid - 1, arr);
+  } else {
+    return binarySearch(val, mid + 1, r, arr);
+  }
+}