Skip to content

41. First Missing Positive

DifficultyHard

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

Thinking

The first idea is a hash set of seen numbers, then probe \(1, 2, 3, \ldots\). Correct and \(O(n)\) time, but \(O(n)\) extra space. The problem asks for \(O(n)\) time and constant extra space; \(n \le 10^5\) makes the set fail the space bound.

The bottleneck is recording whether \(1..n\) appear with extra memory. The missing positive must lie in \([1, n+1]\), so we only care about \(1..n\) — the array indices themselves can be that table.

Swap value \(x\) to index \(x-1\) when \(x \in [1, n]\). In-place swap turns the array into a hash; one more scan finds the first hole.

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 \(O(1)\).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution:
    def firstMissingPositive(self, nums: List[int]) -> int:
        n = len(nums)
        for i in range(n):
            while 1 <= nums[i] <= n and nums[i] != nums[nums[i] - 1]:
                j = nums[i] - 1
                nums[i], nums[j] = nums[j], nums[i]
        for i in range(n):
            if nums[i] != i + 1:
                return i + 1
        return n + 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution {
    public int firstMissingPositive(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            while (nums[i] > 0 && nums[i] <= n && nums[i] != nums[nums[i] - 1]) {
                swap(nums, i, nums[i] - 1);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] != i + 1) {
                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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            while (nums[i] > 0 && nums[i] <= n && nums[i] != nums[nums[i] - 1]) {
                swap(nums[i], nums[nums[i] - 1]);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }
        return n + 1;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
func firstMissingPositive(nums []int) int {
    n := len(nums)
    for i := range nums {
        for 0 < nums[i] && nums[i] <= n && nums[i] != nums[nums[i]-1] {
            nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]
        }
    }
    for i, x := range nums {
        if x != i+1 {
            return i + 1
        }
    }
    return n + 1
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
function firstMissingPositive(nums: number[]): number {
    const n = nums.length;
    for (let i = 0; i < n; i++) {
        while (nums[i] >= 1 && nums[i] <= n && nums[i] !== nums[nums[i] - 1]) {
            const j = nums[i] - 1;
            [nums[i], nums[j]] = [nums[j], nums[i]];
        }
    }
    for (let i = 0; i < n; i++) {
        if (nums[i] !== i + 1) {
            return i + 1;
        }
    }
    return n + 1;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
impl Solution {
    pub fn first_missing_positive(mut nums: Vec<i32>) -> i32 {
        let n = nums.len();
        for i in 0..n {
            while nums[i] > 0 && nums[i] <= n as i32 && nums[i] != nums[nums[i] as usize - 1] {
                let j = nums[i] as usize - 1;
                nums.swap(i, j);
            }
        }
        for i in 0..n {
            if nums[i] != (i + 1) as i32 {
                return (i + 1) as i32;
            }
        }
        return (n + 1) as i32;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
public 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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
int firstMissingPositive(int* nums, int numsSize) {
    for (int i = 0; i < numsSize; ++i) {
        while (nums[i] > 0 && nums[i] <= numsSize && nums[i] != nums[nums[i] - 1]) {
            int j = nums[i] - 1;
            int t = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    }
    for (int i = 0; i < numsSize; ++i) {
        if (nums[i] != i + 1) {
            return i + 1;
        }
    }
    return numsSize + 1;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function firstMissingPositive($nums) {
        $n = count($nums);
        for ($i = 0; $i < $n; $i++) {
            while ($nums[$i] >= 1 && $nums[$i] <= $n && $nums[$i] != $nums[$nums[$i] - 1]) {
                $j = $nums[$i] - 1;
                $t = $nums[$i];
                $nums[$i] = $nums[$j];
                $nums[$j] = $t;
            }
        }
        for ($i = 0; $i < $n; $i++) {
            if ($nums[$i] != $i + 1) {
                return $i + 1;
            }
        }
        return $n + 1;
    }
}

Comments