Given an unsorted integer array nums, find the smallest missing positive integer.
Follow up: Could you 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:
0 <= nums.length <= 300-231 <= nums[i] <= 231 - 1
Solution
cue words from question
unsorted : means that we cannot use binary search Positive intgers : means we do not care about negative numbers as well as duplicates
So from above we conclude that whatever we do needs to be inplace.
Idea 1 : Just put the numbers in their position ex: num[i] should be at num[num[i]]
this fails because if array size is less than num[i] then we get seg fault;
Idea 2: get max and min from array in range [1,max] now we subtract min from every number such that our range becomes >= our array size now we can place our numbers in their position ex : num[i] should be at num[num[i]-min]
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class Solution {
public:
int firstMissingPositive(vector<int>& nums) {
if(nums.size()==0)
return 1;
if(nums.size()==1)
return nums[0]!=1?1:2;
int i=0;
int min_a=INT_MAX,max_a=INT_MIN;
for(int i=0;i<nums.size();i++)
{
if(nums[i]>0)
{min_a=min(min_a,nums[i]);
max_a = max(max_a,nums[i]);}
}
if(min_a!=1)
return 1;
while(i<nums.size())
{
if(nums[i]>0 && nums[i]-min_a < i && nums[nums[i]-min_a]!=nums[i])
{
swap(nums[nums[i]-min_a],nums[i]);
}
else
i++;
}
int ans=1;
for(int i=0;i<nums.size();i++)
if(ans == nums[i])
ans++;
return ans;
}
};