/*
41. First Missing Positive
Hard - 41.8%

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 O(1) auxiliary space.

Example 1:
Input: nums = [1,2,0]
Output: 3
Explanation: The numbers in the range [1,2] are all in the array.

Example 2:
Input: nums = [3,4,-1,1]
Output: 2
Explanation: 1 is in the array but 2 is missing.

Example 3:
Input: nums = [7,8,9,11,12]
Output: 1
Explanation: The smallest positive integer 1 is missing.
*/

class Solution {
    public int firstMissingPositive(int[] nums) {
        int n = nums.length;
        
        // Step 1: Replace all numbers <= 0 or > n with n+1
        // Since we're looking for the first missing positive in range [1, n+1]
        for (int i = 0; i < n; i++) {
            if (nums[i] <= 0 || nums[i] > n) {
                nums[i] = n + 1;
            }
        }
        
        // Step 2: Use array indices as hash map
        // For each number x in [1, n], mark nums[x-1] as negative
        for (int i = 0; i < n; i++) {
            int num = Math.abs(nums[i]);
            if (num <= n) {
                nums[num - 1] = -Math.abs(nums[num - 1]);
            }
        }
        
        // Step 3: Find first positive number
        // The index of first positive number + 1 is our answer
        for (int i = 0; i < n; i++) {
            if (nums[i] > 0) {
                return i + 1;
            }
        }
        
        // If all numbers [1, n] are present, return n+1
        return n + 1;
    }
}