Physical Address

304 North Cardinal St.
Dorchester Center, MA 02124

# First Missing Positive LeetCode Solution

## Problem – First Missing Positive

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
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`

### First Missing Positive LeetCode Solution in C++

``````class Solution
{
public:
int firstMissingPositive(int A[], int n)
{
for(int i = 0; i < n; ++ i)
while(A[i] > 0 && A[i] <= n && A[A[i] - 1] != A[i])
swap(A[i], A[A[i] - 1]);

for(int i = 0; i < n; ++ i)
if(A[i] != i + 1)
return i + 1;

return n + 1;
}
};
``````

### First Missing Positive LeetCode Solution in Python

``````def firstMissingPositive(self, nums):
"""
:type nums: List[int]
:rtype: int
Basic idea:
1. for any array whose length is l, the first missing positive must be in range [1,...,l+1],
so we only have to care about those elements in this range and remove the rest.
2. we can use the array index as the hash to restore the frequency of each number within
the range [1,...,l+1]
"""
nums.append(0)
n = len(nums)
for i in range(len(nums)): #delete those useless elements
if nums[i]<0 or nums[i]>=n:
nums[i]=0
for i in range(len(nums)): #use the index as the hash to record the frequency of each number
nums[nums[i]%n]+=n
for i in range(1,len(nums)):
if nums[i]/n==0:
return i
return n
``````

### First Missing Positive LeetCode Solution in Java

``````public class Solution {
public int firstMissingPositive(int[] A) {
int i = 0;
while(i < A.length){
if(A[i] == i+1 || A[i] <= 0 || A[i] > A.length) i++;
else if(A[A[i]-1] != A[i]) swap(A, i, A[i]-1);
else i++;
}
i = 0;
while(i < A.length && A[i] == i+1) i++;
return i+1;
}

private void swap(int[] A, int i, int j){
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
}
``````
##### First Missing Positive LeetCode Solution Review:

In our experience, we suggest you solve this First Missing Positive LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the First Missing Positive LeetCode Solution

Find on LeetCode

##### Conclusion:

I hope this First Missing Positive LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions