Add solution #1007

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,083 LeetCode solutions in JavaScript
+# 1,084 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -815,6 +815,7 @@
 1004|[Max Consecutive Ones III](./solutions/1004-max-consecutive-ones-iii.js)|Medium|
 1005|[Maximize Sum Of Array After K Negations](./solutions/1005-maximize-sum-of-array-after-k-negations.js)|Easy|
 1006|[Clumsy Factorial](./solutions/1006-clumsy-factorial.js)|Medium|
+1007|[Minimum Domino Rotations For Equal Row](./solutions/1007-minimum-domino-rotations-for-equal-row.js)|Medium|
 1009|[Complement of Base 10 Integer](./solutions/1009-complement-of-base-10-integer.js)|Easy|
 1010|[Pairs of Songs With Total Durations Divisible by 60](./solutions/1010-pairs-of-songs-with-total-durations-divisible-by-60.js)|Medium|
 1022|[Sum of Root To Leaf Binary Numbers](./solutions/1022-sum-of-root-to-leaf-binary-numbers.js)|Easy|

solutions/1007-minimum-domino-rotations-for-equal-row.js

@@ -0,0 +1,40 @@
+/**
+ * 1007. Minimum Domino Rotations For Equal Row
+ * https://leetcode.com/problems/minimum-domino-rotations-for-equal-row/
+ * Difficulty: Medium
+ *
+ * In a row of dominoes, tops[i] and bottoms[i] represent the top and bottom halves of the
+ * ith domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)
+ *
+ * We may rotate the ith domino, so that tops[i] and bottoms[i] swap values.
+ *
+ * Return the minimum number of rotations so that all the values in tops are the same, or all
+ * the values in bottoms are the same.
+ *
+ * If it cannot be done, return -1.
+ */
+
+/**
+ * @param {number[]} tops
+ * @param {number[]} bottoms
+ * @return {number}
+ */
+var minDominoRotations = function(tops, bottoms) {
+  const topResult = countRotations(tops[0]);
+  if (topResult !== -1) return topResult;
+
+  return countRotations(bottoms[0]);
+
+  function countRotations(target) {
+    let topCount = 0;
+    let bottomCount = 0;
+
+    for (let i = 0; i < tops.length; i++) {
+      if (tops[i] !== target && bottoms[i] !== target) return -1;
+      topCount += tops[i] !== target ? 1 : 0;
+      bottomCount += bottoms[i] !== target ? 1 : 0;
+    }
+
+    return Math.min(topCount, bottomCount);
+  }
+};