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In this post, you will find the solution for the First Missing Positive in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.
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Link for the Problem – First Missing Positive – LeetCode Problem
First Missing Positive– LeetCode Problem
Problem:
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 constant extra space.
Example 1:
Input: nums = [1,2,0] Output: 3
Example 2:
Input: nums = [3,4,-1,1] Output: 2
Example 3:
Input: nums = [7,8,9,11,12] Output: 1
Constraints:
1 <= nums.length <= 5 * 105
-231 <= nums[i] <= 231 - 1
First Missing Positive– LeetCode Solutions
class Solution { public: int firstMissingPositive(vector<int>& nums) { const int n = nums.size(); // Correct slot: // nums[i] = i + 1 // nums[i] - 1 = i // nums[nums[i] - 1] = nums[i] 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; } };
class Solution { public int firstMissingPositive(int[] nums) { final int n = nums.length; // Correct slot: // nums[i] = i + 1 // nums[i] - 1 = i // nums[nums[i] - 1] = nums[i] 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) { final int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; } }
class Solution: def firstMissingPositive(self, nums: List[int]) -> int: n = len(nums) # Correct slot: # nums[i] = i + 1 # nums[i] - 1 = i # nums[nums[i] - 1] = nums[i] for i in range(n): while nums[i] > 0 and nums[i] <= n and nums[nums[i] - 1] != nums[i]: nums[nums[i] - 1], nums[i] = nums[i], nums[nums[i] - 1] for i, num in enumerate(nums): if num != i + 1: return i + 1 return n + 1
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