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apps/docs/content/array/268-missing-number.mdx

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+---
+title: '268. Missing Number'
+description: Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array
+sidebar:
+  label: 'Missing Number'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
+::::
+
+### Example 1:
+- Input: `nums = [3,0,1]`
+- Output: `2`
+- Explanation: `n = 3` since there are `3` numbers, so all numbers are in the range `[0,3]`. `2` is the missing number in the range since it does not appear in `nums`.
+
+### Example 2:
+- Input: `nums = [0,1]`
+- Output: `2`
+- Explanation: `n = 2` since there are `2` numbers, so all numbers are in the range `[0,2]`. `2` is the missing number in the range since it does not appear in `nums`.
+
+### Example 3:
+- Input: `nums = [9,6,4,2,3,5,7,0,1]`
+- Output: `8`
+- Explanation: `n = 9` since there are `9` numbers, so all numbers are in the range `[0,9]`. `8` is the missing number in the range since it does not appear in `nums`.
+
+### Constraints:
+
+- `n == nums.length`
+- `1 <= n <= 10^4`
+- `0 <= nums[i] <= n`
+- All the numbers of `nums` are unique.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["missingNumber(nums)"]) --> I["numSet = Set(nums), expectedCount = n + 1"]
+  I --> L{"more i in 0..expectedCount-1?"}
+  L -- yes --> Q{"numSet has i?"}
+  Q -- no --> R(["return i"])
+  Q -- yes --> L
+  L -- no --> E(["return -1"])
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var missingNumber = function(nums) {
+  const numSet = new Set(nums);
+  const expectedCount = nums.length + 1;
+
+  for (let i = 0; i < expectedCount; i++){
+    if(!numSet.has(i)){
+      return i;
+    }
+  }
+  return -1;
+};
+```
+
+## Explanation
+
+<YouTube id="WnPLSRLSANE" title="Missing Number - Blind 75 - Leetcode 268 - Python" />