class Solution { public int firstMissingPositive(int[] nums) { Set<Integer> set = new HashSet<Integer>(); for(int num : nums) set.add(num); int i = 1; for(; i <= nums.length; i++) { if(!set.contains(i)) return i; } return i; } }

class Solution { public int firstMissingPositive(int[] nums) { if(nums == null || nums.length == 0) return 1; int n = nums.length; boolean containsOne = false; // Step 1 : Sets the containsOne flag to true if input array contains 1 , // also sets all the negative numbers, zero and number's greater than n, to 1  // Make sure you don't mess up the order of statements for(int i = 0; i < n; i++) { if(nums[i] == 1) containsOne = true; else if(nums[i] <= 0 || nums[i] > n) nums[i] = 1; } // if array doesn't contains 1, the first missing positive will be 1 if(!containsOne) return 1; // Step 2 : This step marks the number we have seen, by setting its index  // position to a negative number  // e.g. if nums[i] = 7, since 7 should come at index 6 (that's why - 1, array indexes starts with 0) // we mark nums[6] = nums[6] * -1 (if nums[6] was not already negative) for(int i = 0; i < n; i++) { int index = Math.abs(nums[i]) - 1; // if its already negative, we don't do anything if(nums[index] > 0) nums[index] *= -1; } // Step 3 : Finally, look for a non negative number in the array // If we find an index i, such that nums[i] is positive, that means // we didn't see the number which should come at this index i // the number that should come at index i should be i+1, so we return i+1 for(int i = 0; i < n; i++) { if(nums[i] > 0) return i + 1; } // If we reach here it means, original input array had all the positive numbers  // from 1 to n, so the first missing positive number will be n + 1 return n + 1; } }