Refactor dynamic programming approach of Trapping Rain Water problem.

Author: trekhleb

src/algorithms/uncategorized/rain-terraces/dpRainTerraces.js

@@ -5,81 +5,41 @@
  * @return {number}
  */
 export default function dpRainTerraces(terraces) {
-  /*
-   * STEPS
-   *
-   * 1. Find the highest terraces on the left and right side of the elevation map:
-   * e.g. for [0, 2, 4, 3, 4, 2, 4, 0, 8, 7, 0] we would have leftMax = 4 and rightMax = 8.
-   * This is because water will "trail off" the sides of the terraces.
-   *
-   * 2. At this point, we are essentially dealing with a new map: [4, 3, 4, 2, 4, 0, 8].
-   * From here, we loop through the map from the left to the right if leftMax < rightMax
-   * (otherwise we move from right to left), adding water as we go unless we reach a value
-   * that is greater than or equal to leftMax or rightMax.
-   * e.g. [4, 3, 4, 2, 4, 0, 8]
-   *             ^
-   * water = water + (leftMax - 3) = 1
-   *
-   * or if the terrace array was reversed:
-   * e.g. [8, 0, 4, 2, 4, 3, 4]
-   *                   ^
-   * water = water + (rightMax - 3) = 1
-   *
-   * 3. Again, we've essentially shortened the map: [4, 2, 4, 0, 8].
-   * Now we repeat the above steps on the new array.
-   *
-   * Next Iteration:
-   * [4, 2, 4, 0, 8]
-   *        ^
-   * water = water + (leftMax - 2) = 3
-   *
-   * Next Iteration:
-   * [4, 0, 8]
-   *        ^
-   * water = water + (leftMax - 0) = 7
-   *
-   * 4. Return result: 7
-   */
-  let leftIndex = 0;
-  let rightIndex = terraces.length - 1;
-
-  let leftMaxLevel = 0;
-  let rightMaxLevel = 0;
-
   let waterAmount = 0;
 
-  while (leftIndex < rightIndex) {
-    // Loop to find the highest terrace from the left side.
-    while (leftIndex < rightIndex && terraces[leftIndex] <= terraces[leftIndex + 1]) {
-      leftIndex += 1;
-    }
-
-    leftMaxLevel = terraces[leftIndex];
-
-    // Loop to find the highest terrace from the right side.
-    while (rightIndex > leftIndex && terraces[rightIndex] <= terraces[rightIndex - 1]) {
-      rightIndex -= 1;
-    }
-
-    rightMaxLevel = terraces[rightIndex];
-
-    // Determine which direction we need to go.
-    if (leftMaxLevel < rightMaxLevel) {
-      // Move from left to right and collect water.
-      leftIndex += 1;
+  // 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],
+    );
+  }
 
-      while (leftIndex < rightIndex && terraces[leftIndex] <= leftMaxLevel) {
-        waterAmount += leftMaxLevel - terraces[leftIndex];
-        leftIndex += 1;
-      }
-    } else {
-      // Move from right to left and collect water.
-      rightIndex -= 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],
+    );
+  }
 
-      while (leftIndex < rightIndex && terraces[rightIndex] <= rightMaxLevel) {
-        waterAmount += rightMaxLevel - terraces[rightIndex];
-        rightIndex -= 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];
     }
   }