Add solution #763

Author: JoshCrozier

README.md

@@ -421,6 +421,7 @@
 747|[Largest Number At Least Twice of Others](./0747-largest-number-at-least-twice-of-others.js)|Easy|
 748|[Shortest Completing Word](./0748-shortest-completing-word.js)|Easy|
 762|[Prime Number of Set Bits in Binary Representation](./0762-prime-number-of-set-bits-in-binary-representation.js)|Easy|
+763|[Partition Labels](./0763-partition-labels.js)|Medium|
 784|[Letter Case Permutation](./0784-letter-case-permutation.js)|Medium|
 790|[Domino and Tromino Tiling](./0790-domino-and-tromino-tiling.js)|Medium|
 791|[Custom Sort String](./0791-custom-sort-string.js)|Medium|

solutions/0763-partition-labels.js

@@ -0,0 +1,37 @@
+/**
+ * 763. Partition Labels
+ * https://leetcode.com/problems/partition-labels/
+ * Difficulty: Medium
+ *
+ * You are given a string s. We want to partition the string into as many parts as
+ * possible so that each letter appears in at most one part. For example, the
+ * string "ababcc" can be partitioned into ["abab", "cc"], but partitions such as
+ * ["aba", "bcc"] or ["ab", "ab", "cc"] are invalid.
+ *
+ * Note that the partition is done so that after concatenating all the parts in
+ * order, the resultant string should be s.
+ *
+ * Return a list of integers representing the size of these parts.
+ */
+
+/**
+ * @param {string} s
+ * @return {number[]}
+ */
+var partitionLabels = function(s) {
+  const map = new Map();
+  for (let i = 0; i < s.length; i++) {
+    map.set(s[i], i);
+  }
+
+  const result = [];
+  for (let i = 0, start = 0, end = 0; i < s.length; i++) {
+    end = Math.max(end, map.get(s[i]));
+    if (i === end) {
+      result.push(end - start + 1);
+      start = i + 1;
+    }
+  }
+
+  return result;
+};