From 812bab6ec0aa5645b8665458ab6bfd70bf4f5bee Mon Sep 17 00:00:00 2001
From: mankalavaishnavi <mankalavaishnavi2004@gmail.com>
Date: Thu, 13 Jun 2024 20:19:34 +0530
Subject: [PATCH] done

---
 .../0000-0099/0041-First-Missing-Positive.md  | 155 ++++++++++++++++++
 1 file changed, 155 insertions(+)
 create mode 100644 dsa-solutions/lc-solutions/0000-0099/0041-First-Missing-Positive.md

diff --git a/dsa-solutions/lc-solutions/0000-0099/0041-First-Missing-Positive.md b/dsa-solutions/lc-solutions/0000-0099/0041-First-Missing-Positive.md
new file mode 100644
index 000000000..0fce1bff4
--- /dev/null
+++ b/dsa-solutions/lc-solutions/0000-0099/0041-First-Missing-Positive.md
@@ -0,0 +1,155 @@
+---
+id: 41FirstMissingPositive
+title: First Missing Positive (LeetCode)
+sidebar_label: 0041-First Missing Positive
+tags:
+  - Array
+  - Hash Table
+description: Find the smallest missing positive integer.
+sidebar_position: 41
+---
+
+## Problem Description
+
+| Problem Statement | Solution Link | LeetCode Profile |
+| :---------------- | :------------ | :--------------- |
+| [First Missing Positive](https://leetcode.com/problems/first-missing-positive/description/) | [First Missing Positive Solution on LeetCode](https://leetcode.com/problems/first-missing-positive/solutions/) | [vaishu_1904](https://leetcode.com/u/vaishu_1904/) |
+
+
+## Problem Description
+
+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.
+
+### Example 1
+
+- **Input:** `nums = [1,2,0]`
+- **Output:** `3`
+- **Explanation:** The smallest missing positive integer is `3`.
+
+### Example 2
+
+- **Input:** `nums = [3,4,-1,1]`
+- **Output:** `2`
+- **Explanation:** The smallest missing positive integer is `2`.
+
+### Example 3
+
+- **Input:** `nums = [7,8,9,11,12]`
+- **Output:** `1`
+- **Explanation:** The smallest missing positive integer is `1`.
+
+### Constraints
+
+- `1 <= nums.length <= 10^5`
+- `-2^31 <= nums[i] <= 2^31 - 1`
+
+## Approach
+
+To solve the problem, we can use the following approach:
+
+1. **Mark Elements Out of Range**:
+   - Iterate through the array and mark elements that are out of the range `[1, n]` by setting them to a number greater than `n`.
+   
+2. **Use Indices as Markers**:
+   - Use the indices of the array to mark the presence of numbers by negating the value at the corresponding index.
+
+3. **Identify the Missing Positive**:
+   - Iterate through the array again to find the first positive value, which indicates the missing positive integer.
+
+### Solution Code
+
+#### Python
+
+```python
+class Solution:
+    def firstMissingPositive(self, nums: List[int]) -> int:
+        n = len(nums)
+
+        for i in range(n):
+            if nums[i] <= 0 or nums[i] > n:
+                nums[i] = n + 1
+
+        for i in range(n):
+            num = abs(nums[i])
+            if num <= n:
+                nums[num - 1] = -abs(nums[num - 1])
+
+        for i in range(n):
+            if nums[i] > 0:
+                return i + 1
+
+        return n + 1
+```
+
+#### Java
+
+```java
+class Solution {
+    public int firstMissingPositive(int[] nums) {
+        int n = nums.length;
+
+        for (int i = 0; i < n; i++) {
+            if (nums[i] <= 0 || nums[i] > n) {
+                nums[i] = n + 1;
+            }
+        }
+
+        for (int i = 0; i < n; i++) {
+            int num = Math.abs(nums[i]);
+            if (num <= n) {
+                nums[num - 1] = -Math.abs(nums[num - 1]);
+            }
+        }
+
+        for (int i = 0; i < n; i++) {
+            if (nums[i] > 0) {
+                return i + 1;
+            }
+        }
+
+        return n + 1;
+    }
+}
+```
+
+#### C++
+
+```c++
+#include <vector>
+#include <cmath>
+
+using namespace std;
+
+class Solution {
+public:
+    int firstMissingPositive(vector<int>& nums) {
+        int n = nums.size();
+
+        for (int i = 0; i < n; i++) {
+            if (nums[i] <= 0 || nums[i] > n) {
+                nums[i] = n + 1;
+            }
+        }
+
+        for (int i = 0; i < n; i++) {
+            int num = abs(nums[i]);
+            if (num <= n) {
+                nums[num - 1] = -abs(nums[num - 1]);
+            }
+        }
+
+        for (int i = 0; i < n; i++) {
+            if (nums[i] > 0) {
+                return i + 1;
+            }
+        }
+
+        return n + 1;
+    }
+};
+```
+
+### Conclusion:
+This approach ensures that the smallest missing positive integer is found efficiently, using constant extra space.
\ No newline at end of file