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41. First Missing Positive
Description
Given an unsorted integer array nums
, return the smallest missing positive integer.
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 { 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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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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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 { if i+1 != v { return i + 1 } } return n + 1 }
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function firstMissingPositive(nums: number[]): number { const n = nums.length; let i = 0; while (i < n) { const j = nums[i] - 1; if (j === i || j < 0 || j >= n || nums[i] === nums[j]) { i++; } else { [nums[i], nums[j]] = [nums[j], nums[i]]; } } const res = nums.findIndex((v, i) => v !== i + 1); return (res === -1 ? n : res) + 1; }
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public class Solution { public int FirstMissingPositive(int[] nums) { var i = 0; while (i < nums.Length) { if (nums[i] > 0 && nums[i] <= nums.Length) { var index = nums[i] -1; if (index != i && nums[index] != nums[i]) { var temp = nums[i]; nums[i] = nums[index]; nums[index] = temp; } else { ++i; } } else { ++i; } } for (i = 0; i < nums.Length; ++i) { if (nums[i] != i + 1) { return i + 1; } } return nums.Length + 1; } }
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impl Solution { pub fn first_missing_positive(mut nums: Vec<i32>) -> i32 { let n = nums.len(); let mut i = 0; while i < n { let j = nums[i] - 1; if (i as i32) == j || j < 0 || j >= (n as i32) || nums[i] == nums[j as usize] { i += 1; } else { nums.swap(i, j as usize); } } ( nums .iter() .enumerate() .position(|(i, &v)| (v as usize) != i + 1) .unwrap_or(n) as i32 ) + 1 } }
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class Solution { /** * @param integer[] $nums * @return integer */ function firstMissingPositive($nums) { $n = count($nums); for ($i = 0; $i < $n; $i++) { if ($nums[$i] <= 0) { $nums[$i] = $n + 1; } } for ($i = 0; $i < $n; $i++) { $num = abs($nums[$i]); if ($num <= $n) { $nums[$num - 1] = -abs($nums[$num - 1]); } } for ($i = 0; $i < $n; $i++) { if ($nums[$i] > 0) { return $i + 1; } } return $n + 1; } }