39. First Missing Positive

Place each number i at index i-1. Then scan for the first index where nums[i] != i+1.

Approach

1. For each nums[i] in range [1, n], swap it to position nums[i]-1.
2. Scan array: first index where nums[i] != i+1 is the answer.
3. If all match, answer is n+1.

Complexity

• Time: O(n)
• Space: O(1)

class Solution {
    public int firstMissingPositive(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            while (nums[i] >= 1 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) {
                int temp = nums[nums[i] - 1];
                nums[nums[i] - 1] = nums[i];
                nums[i] = temp;
            }
        }
        for (int i = 0; i < n; i++) {
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }
        return n + 1;
    }
}