From 680ee713a6205544a732c0cb6527956957df4f33 Mon Sep 17 00:00:00 2001
From: Bruce Feldman <bfeldman@sunriselabs.com>
Date: Fri, 10 Aug 2018 13:26:37 -0400
Subject: [PATCH] Make Rain Terraces more terse using standard array functions

---
 .../rain-terraces/dpRainTerraces.js           | 44 +++----------------
 1 file changed, 6 insertions(+), 38 deletions(-)

diff --git a/src/algorithms/uncategorized/rain-terraces/dpRainTerraces.js b/src/algorithms/uncategorized/rain-terraces/dpRainTerraces.js
index 1c48cbd8..9c4f9ad5 100644
--- a/src/algorithms/uncategorized/rain-terraces/dpRainTerraces.js
+++ b/src/algorithms/uncategorized/rain-terraces/dpRainTerraces.js
@@ -5,43 +5,11 @@
  * @return {number}
  */
 export default function dpRainTerraces(terraces) {
-  let waterAmount = 0;
+  // Obtain the maximum elevations to the left and right of each terrace (inclusive).
+  const maxL = terraces.reduce((acc, val, idx) => [...acc, Math.max(val, (acc[idx - 1] || 0))], []);
+  const maxR = terraces.reduceRight((acc, val) => [Math.max(val, (acc[0] || 0)), ...acc], []);
 
-  // Init arrays that will keep the list of left and right maximum levels for specific positions.
-  const leftMaxLevels = new Array(terraces.length).fill(0);
-  const rightMaxLevels = new Array(terraces.length).fill(0);
-
-  // Calculate the highest terrace level from the LEFT relative to the current terrace.
-  [leftMaxLevels[0]] = terraces;
-  for (let terraceIndex = 1; terraceIndex < terraces.length; terraceIndex += 1) {
-    leftMaxLevels[terraceIndex] = Math.max(
-      terraces[terraceIndex],
-      leftMaxLevels[terraceIndex - 1],
-    );
-  }
-
-  // Calculate the highest terrace level from the RIGHT relative to the current terrace.
-  rightMaxLevels[terraces.length - 1] = terraces[terraces.length - 1];
-  for (let terraceIndex = terraces.length - 2; terraceIndex >= 0; terraceIndex -= 1) {
-    rightMaxLevels[terraceIndex] = Math.max(
-      terraces[terraceIndex],
-      rightMaxLevels[terraceIndex + 1],
-    );
-  }
-
-  // Not let's go through all terraces one by one and calculate how much water
-  // each terrace may accumulate based on previously calculated values.
-  for (let terraceIndex = 0; terraceIndex < terraces.length; terraceIndex += 1) {
-    // Pick the lowest from the left/right highest terraces.
-    const currentTerraceBoundary = Math.min(
-      leftMaxLevels[terraceIndex],
-      rightMaxLevels[terraceIndex],
-    );
-
-    if (currentTerraceBoundary > terraces[terraceIndex]) {
-      waterAmount += currentTerraceBoundary - terraces[terraceIndex];
-    }
-  }
-
-  return waterAmount;
+  // Convert maximum values to amount of water at current index and sum water values.
+  const trap = terraces.map((val, idx) => Math.max(0, Math.min(maxL[idx], maxR[idx]) - val));
+  return trap.reduce((acc, val) => acc + val);
 }