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Refactor dynamic programming approach of Trapping Rain Water problem.
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* @return {number}
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* @return {number}
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*/
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*/
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export default function dpRainTerraces(terraces) {
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export default function dpRainTerraces(terraces) {
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/*
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* STEPS
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*
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* 1. Find the highest terraces on the left and right side of the elevation map:
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* e.g. for [0, 2, 4, 3, 4, 2, 4, 0, 8, 7, 0] we would have leftMax = 4 and rightMax = 8.
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* This is because water will "trail off" the sides of the terraces.
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*
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* 2. At this point, we are essentially dealing with a new map: [4, 3, 4, 2, 4, 0, 8].
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* From here, we loop through the map from the left to the right if leftMax < rightMax
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* (otherwise we move from right to left), adding water as we go unless we reach a value
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* that is greater than or equal to leftMax or rightMax.
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* e.g. [4, 3, 4, 2, 4, 0, 8]
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* ^
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* water = water + (leftMax - 3) = 1
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*
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* or if the terrace array was reversed:
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* e.g. [8, 0, 4, 2, 4, 3, 4]
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* ^
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* water = water + (rightMax - 3) = 1
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*
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* 3. Again, we've essentially shortened the map: [4, 2, 4, 0, 8].
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* Now we repeat the above steps on the new array.
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*
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* Next Iteration:
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* [4, 2, 4, 0, 8]
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* ^
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* water = water + (leftMax - 2) = 3
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*
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* Next Iteration:
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* [4, 0, 8]
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* ^
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* water = water + (leftMax - 0) = 7
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*
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* 4. Return result: 7
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*/
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let leftIndex = 0;
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let rightIndex = terraces.length - 1;
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let leftMaxLevel = 0;
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let rightMaxLevel = 0;
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let waterAmount = 0;
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let waterAmount = 0;
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while (leftIndex < rightIndex) {
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// Init arrays that will keep the list of left and right maximum levels for specific positions.
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// Loop to find the highest terrace from the left side.
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const leftMaxLevels = new Array(terraces.length).fill(0);
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while (leftIndex < rightIndex && terraces[leftIndex] <= terraces[leftIndex + 1]) {
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const rightMaxLevels = new Array(terraces.length).fill(0);
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leftIndex += 1;
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}
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leftMaxLevel = terraces[leftIndex];
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// Calculate the highest terrace level from the LEFT relative to the current terrace.
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[leftMaxLevels[0]] = terraces;
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for (let terraceIndex = 1; terraceIndex < terraces.length; terraceIndex += 1) {
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leftMaxLevels[terraceIndex] = Math.max(
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terraces[terraceIndex],
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leftMaxLevels[terraceIndex - 1],
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);
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}
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// Loop to find the highest terrace from the right side.
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// Calculate the highest terrace level from the RIGHT relative to the current terrace.
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while (rightIndex > leftIndex && terraces[rightIndex] <= terraces[rightIndex - 1]) {
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rightMaxLevels[terraces.length - 1] = terraces[terraces.length - 1];
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rightIndex -= 1;
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for (let terraceIndex = terraces.length - 2; terraceIndex >= 0; terraceIndex -= 1) {
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}
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rightMaxLevels[terraceIndex] = Math.max(
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terraces[terraceIndex],
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rightMaxLevels[terraceIndex + 1],
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);
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}
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rightMaxLevel = terraces[rightIndex];
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// Not let's go through all terraces one by one and calculate how much water
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// each terrace may accumulate based on previously calculated values.
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for (let terraceIndex = 0; terraceIndex < terraces.length; terraceIndex += 1) {
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// Pick the lowest from the left/right highest terraces.
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const currentTerraceBoundary = Math.min(
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leftMaxLevels[terraceIndex],
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rightMaxLevels[terraceIndex],
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);
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// Determine which direction we need to go.
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if (currentTerraceBoundary > terraces[terraceIndex]) {
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if (leftMaxLevel < rightMaxLevel) {
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waterAmount += currentTerraceBoundary - terraces[terraceIndex];
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// Move from left to right and collect water.
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leftIndex += 1;
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while (leftIndex < rightIndex && terraces[leftIndex] <= leftMaxLevel) {
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waterAmount += leftMaxLevel - terraces[leftIndex];
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leftIndex += 1;
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}
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} else {
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// Move from right to left and collect water.
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rightIndex -= 1;
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while (leftIndex < rightIndex && terraces[rightIndex] <= rightMaxLevel) {
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waterAmount += rightMaxLevel - terraces[rightIndex];
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rightIndex -= 1;
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}
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}
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}
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}
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}
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