Add solution #758

Author: JoshCrozier

README.md

@@ -687,6 +687,7 @@
 754|[Reach a Number](./solutions/0754-reach-a-number.js)|Medium|
 756|[Pyramid Transition Matrix](./solutions/0756-pyramid-transition-matrix.js)|Medium|
 757|[Set Intersection Size At Least Two](./solutions/0757-set-intersection-size-at-least-two.js)|Hard|
+758|[Bold Words in String](./solutions/0758-bold-words-in-string.js)|Medium|
 761|[Special Binary String](./solutions/0761-special-binary-string.js)|Hard|
 762|[Prime Number of Set Bits in Binary Representation](./solutions/0762-prime-number-of-set-bits-in-binary-representation.js)|Easy|
 763|[Partition Labels](./solutions/0763-partition-labels.js)|Medium|

solutions/0758-bold-words-in-string.js

@@ -0,0 +1,57 @@
+/**
+ * 758. Bold Words in String
+ * https://leetcode.com/problems/bold-words-in-string/
+ * Difficulty: Medium
+ *
+ * Given an array of keywords words and a string s, make all appearances of all keywords words[i]
+ * in s bold. Any letters between <b> and </b> tags become bold.
+ *
+ * Return s after adding the bold tags. The returned string should use the least number of tags
+ * possible, and the tags should form a valid combination.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {string} s
+ * @return {string}
+ */
+var boldWords = function(words, s) {
+  const boldIntervals = [];
+
+  for (const word of words) {
+    let start = s.indexOf(word);
+    while (start !== -1) {
+      boldIntervals.push([start, start + word.length]);
+      start = s.indexOf(word, start + 1);
+    }
+  }
+
+  if (!boldIntervals.length) return s;
+
+  boldIntervals.sort((a, b) => a[0] - b[0] || a[1] - b[1]);
+
+  const mergedIntervals = [];
+  let [currentStart, currentEnd] = boldIntervals[0];
+
+  for (let i = 1; i < boldIntervals.length; i++) {
+    const [nextStart, nextEnd] = boldIntervals[i];
+    if (nextStart <= currentEnd) {
+      currentEnd = Math.max(currentEnd, nextEnd);
+    } else {
+      mergedIntervals.push([currentStart, currentEnd]);
+      [currentStart, currentEnd] = [nextStart, nextEnd];
+    }
+  }
+  mergedIntervals.push([currentStart, currentEnd]);
+
+  let result = '';
+  let lastEnd = 0;
+  for (const [start, end] of mergedIntervals) {
+    result += s.slice(lastEnd, start) + '<b>' + s.slice(start, end) + '</b>';
+    lastEnd = end;
+  }
+
+  result += s.slice(lastEnd);
+
+  return result;
+};