Add solution #41

Author: JoshCrozier

README.md

@@ -17,6 +17,7 @@
 27|[Remove Element](./0027-remove-element.js)|Easy|
 31|[Next Permutation](./0031-next-permutation.js)|Medium|
 36|[Valid Sudoku](./0036-valid-sudoku.js)|Medium|
+41|[First Missing Positive](./0041-first-missing-positive.js)|Hard|
 43|[Multiply Strings](./0043-multiply-strings.js)|Medium|
 54|[Spiral Matrix](./0054-spiral-matrix.js)|Medium|
 58|[Length of Last Word](./0058-length-of-last-word.js)|Easy|

solutions/0041-first-missing-positive.js

@@ -0,0 +1,33 @@
+/**
+ * 41. First Missing Positive
+ * https://leetcode.com/problems/first-missing-positive/
+ * Difficulty: Hard
+ *
+ * Given an unsorted integer array nums, return the smallest missing positive integer.
+ *
+ * You must implement an algorithm that runs in O(n) time and uses constant extra space.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var firstMissingPositive = function(nums) {
+  let i = 0;
+
+  while (i < nums.length) {
+    if (nums[i] > 0 && nums[i] <= nums.length && nums[nums[i] - 1] !== nums[i]) {
+      [nums[nums[i] - 1], nums[i]] = [nums[i], nums[nums[i] - 1]];
+    } else {
+      i++;
+    }
+  }
+
+  for (i = 0; i < nums.length; i++) {
+    if (nums[i] !== i + 1) {
+      return i + 1;
+    }
+  }
+
+  return i + 1;
+};