Add solution #1170

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,163 LeetCode solutions in JavaScript
+# 1,164 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -911,6 +911,7 @@
 1162|[As Far from Land as Possible](./solutions/1162-as-far-from-land-as-possible.js)|Medium|
 1163|[Last Substring in Lexicographical Order](./solutions/1163-last-substring-in-lexicographical-order.js)|Hard|
 1169|[Invalid Transactions](./solutions/1169-invalid-transactions.js)|Medium|
+1170|[Compare Strings by Frequency of the Smallest Character](./solutions/1170-compare-strings-by-frequency-of-the-smallest-character.js)|Medium|
 1189|[Maximum Number of Balloons](./solutions/1189-maximum-number-of-balloons.js)|Easy|
 1200|[Minimum Absolute Difference](./solutions/1200-minimum-absolute-difference.js)|Easy|
 1206|[Design Skiplist](./solutions/1206-design-skiplist.js)|Hard|

solutions/1170-compare-strings-by-frequency-of-the-smallest-character.js

@@ -0,0 +1,55 @@
+/**
+ * 1170. Compare Strings by Frequency of the Smallest Character
+ * https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/
+ * Difficulty: Medium
+ *
+ * Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty
+ * string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest
+ * character is 'c', which has a frequency of 2.
+ *
+ * You are given an array of strings words and another array of query strings queries. For each
+ * query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each
+ * W in words.
+ *
+ * Return an integer array answer, where each answer[i] is the answer to the ith query.
+ */
+
+/**
+ * @param {string[]} queries
+ * @param {string[]} words
+ * @return {number[]}
+ */
+var numSmallerByFrequency = function(queries, words) {
+  const wordFrequencies = words.map(getMinCharFreq).sort((a, b) => a - b);
+
+  return queries.map(query => {
+    const queryFreq = getMinCharFreq(query);
+    let left = 0;
+    let right = wordFrequencies.length;
+
+    while (left < right) {
+      const mid = Math.floor((left + right) / 2);
+      if (wordFrequencies[mid] <= queryFreq) {
+        left = mid + 1;
+      } else {
+        right = mid;
+      }
+    }
+
+    return wordFrequencies.length - left;
+  });
+
+  function getMinCharFreq(str) {
+    let minChar = 'z';
+    let freq = 0;
+    for (const char of str) {
+      if (char < minChar) {
+        minChar = char;
+        freq = 1;
+      } else if (char === minChar) {
+        freq++;
+      }
+    }
+    return freq;
+  }
+};