First Missing Positive

Problem

Given an unsorted integer array nums. Return the smallest positive integer that is not present in nums.

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.

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1

Solution

Another Cyclic Sort

class Solution {
    public int firstMissingPositive(int[] nums) {
        var i = 0;
        while (i < nums.length) {
            if (nums[i] > 0 && nums[i] <= nums.length && nums[i] != nums[nums[i] - 1]) {
                swap(nums, i, nums[i] - 1);
            } else {
                i += 1;
            }
        }

        for (i = 0; i < nums.length; i++) {
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }

        return nums.length + 1;
    }

    void swap(int[] nums, int l, int r) {
        nums[l] ^= nums[r];
        nums[r] ^= nums[l];
        nums[l] ^= nums[r];
    }
}

Cyclic Sort

class Solution {
    public int firstMissingPositive(int[] nums) {
        // cyclic sort
        var i = 0;
        while (i < nums.length) {
            var dst = nums[i] - 1;
            if (nums[i] > 0 && nums[i] <= nums.length && nums[i] != nums[dst]) {
                swap(nums, i, dst);
            } else {
                i += 1;
            }
        }

        for (i = 0; i < nums.length; i++) {
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }

        return nums.length + 1;
    }

    void swap(int[] nums, int x, int y) {
        var tmp = nums[x];
        nums[x] = nums[y];
        nums[y] = tmp;
    }
}

XOR

This didn’t work because the array allows duplicate numbers :/

class Solution {
    public int firstMissingPositive(int[] nums) {
        var x = 0;
        var min = Integer.MAX_VALUE;
        var max = Integer.MIN_VALUE;
        for (var n : nums) {
            if (n < 1) {
                continue;
            }
            min = Math.min(min, n);
            max = Math.max(max, n);
            x ^= n;
        }
        if (min > 1) {
            return 1;
        }
        for (var i = min; i <= max; i++) {
            x ^= i;
        }
        if (x == 0) {
            return max + 1;
        }
        return x;
    }
}