Add solution #2158

Author: JoshCrozier

README.md

@@ -1951,6 +1951,7 @@
 2155|[All Divisions With the Highest Score of a Binary Array](./solutions/2155-all-divisions-with-the-highest-score-of-a-binary-array.js)|Medium|
 2156|[Find Substring With Given Hash Value](./solutions/2156-find-substring-with-given-hash-value.js)|Hard|
 2157|[Groups of Strings](./solutions/2157-groups-of-strings.js)|Hard|
+2158|[Amount of New Area Painted Each Day](./solutions/2158-amount-of-new-area-painted-each-day.js)|Hard|
 2160|[Minimum Sum of Four Digit Number After Splitting Digits](./solutions/2160-minimum-sum-of-four-digit-number-after-splitting-digits.js)|Easy|
 2161|[Partition Array According to Given Pivot](./solutions/2161-partition-array-according-to-given-pivot.js)|Medium|
 2162|[Minimum Cost to Set Cooking Time](./solutions/2162-minimum-cost-to-set-cooking-time.js)|Medium|

solutions/2158-amount-of-new-area-painted-each-day.js

@@ -0,0 +1,65 @@
+/**
+ * 2158. Amount of New Area Painted Each Day
+ * https://leetcode.com/problems/amount-of-new-area-painted-each-day/
+ * Difficulty: Hard
+ *
+ * There is a long and thin painting that can be represented by a number line. You are given
+ * a 0-indexed 2D integer array paint of length n, where paint[i] = [starti, endi]. This means
+ * that on the ith day you need to paint the area between starti and endi.
+ *
+ * Painting the same area multiple times will create an uneven painting so you only want to
+ * paint each area of the painting at most once.
+ *
+ * Return an integer array worklog of length n, where worklog[i] is the amount of new area that
+ * you painted on the ith day.
+ */
+
+/**
+ * @param {number[][]} paint
+ * @return {number[]}
+ */
+var amountPainted = function(paint) {
+ const intervals = new Map();
+ const result = [];
+
+ for (const [start, end] of paint) {
+ let left = start;
+ let right = end;
+ let paintAmount = right - left;
+
+ const keys = Array.from(intervals.keys()).sort((a, b) => a - b);
+ const toRemove = [];
+ let merged = false;
+
+ for (const key of keys) {
+ const value = intervals.get(key);
+
+ if (key > right || value < left) continue;
+
+ if (!merged && key <= left && value >= left) {
+ left = Math.min(left, key);
+ right = Math.max(right, value);
+ paintAmount = Math.max(0, end - Math.max(start, value));
+ toRemove.push(key);
+ merged = true;
+ } else if (key >= left && key < right) {
+ paintAmount -= Math.min(right, value) - key;
+ right = Math.max(right, value);
+ toRemove.push(key);
+ } else if (value > left && value <= right) {
+ paintAmount -= value - Math.max(left, key);
+ left = Math.min(left, key);
+ toRemove.push(key);
+ }
+ }
+
+ for (const key of toRemove) {
+ intervals.delete(key);
+ }
+
+ intervals.set(left, right);
+ result.push(Math.max(0, paintAmount));
+ }
+
+ return result;
+};