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41. First Missing Positive

Description

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

Solutions

Solution 1: In-place Swap

We assume the length of the array $nums$ is $n$, then the smallest positive integer must be in the range $[1, .., n + 1]$. We can traverse the array and swap each number $x$ to its correct position, that is, the position $x - 1$. If $x$ is not in the range $[1, n + 1]$, then we can ignore it.

After the traversal, we traverse the array again. If $i+1$ is not equal to $nums[i]$, then $i+1$ is the smallest positive integer we are looking for.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

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class Solution:
    def firstMissingPositive(self, nums: List[int]) -> int:
        def swap(i, j):
            nums[i], nums[j] = nums[j], nums[i]

        n = len(nums)
        for i in range(n):
            while 1 <= nums[i] <= n and nums[i] != nums[nums[i] - 1]:
                swap(i, nums[i] - 1)
        for i in range(n):
            if i + 1 != nums[i]:
                return i + 1
        return n + 1

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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[i] != nums[nums[i] - 1]) {
                swap(nums, i, nums[i] - 1);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (i + 1 != nums[i]) {
                return i + 1;
            }
        }
        return n + 1;
    }

    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

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

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func firstMissingPositive(nums []int) int {
    n := len(nums)
    for i := range nums {
        for nums[i] >= 1 && nums[i] <= n && nums[i] != nums[nums[i]-1] {
            nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]
        }
    }
    for i, v := range nums {
        <