# Missing Positive Integer

• Algo:

1. Separate all positive numbers to the right and move forward to the problem.
2. Find the missing positive the positive elements by :
• Mark arr[i] as visited by making arr[arr[i] - 1] negative.
... - Return the first index value at which is positive

/* swap two integers */
void swap(int *a, int *b)
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}

/* puts all non-positive (0 and negative) numbers on left
side of arr[] and return count of such numbers */
int separateNP(int arr[], int size)
{
int j = 0, i;
for(i = 0; i < size; i++)
{
if (arr[i] <= 0)
{
swap(&arr[i], &arr[j]);
j++; // increment count of non-positive integers
}
}

``````return j;
``````

}

int getMissing(int arr[], int size)
{
int i;

// Mark arr[i] as visited by making arr[arr[i] - 1] negative. Note that
// 1 is subtracted because index start from 0 and positive numbers start from 1
for(i = 0; i < size; i++)
{
if(abs(arr[i]) - 1 < size && arr[abs(arr[i]) - 1] > 0)
arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
}

// Return the first index value at which is positive
for(i = 0; i < size; i++)
if (arr[i] > 0)
return i+1; // 1 is added becuase indexes start from 0

return size+1;
}

int firstMissingPositive(int* nums, int numsSize) {
// First separate positive and negative numbers
int shift = separateNP (nums,numsSize );

// Shift the array and call findMissingPositive for
// positive part
return getMissing(nums+shift, numsSize-shift);
}
...

• Python3 solution

Algo:

• convert the list into a set(finding an element in set is O(1))
• iterate from 1 to (1+"length of nums"+1) and first number we can't find is the answer

Why (1+"length of nums"+1):

• we are to find first missing positive number(which start from 1)
• if "nums" start with 1: maximum value it can go to is 1+"length of nums"
• all other cases, we will for sure find missing number before: 1+"length of nums" value
``````class Solution:
def firstMissingPositive(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 0:
return 1

nums = set(nums)

for i in range(1, 1+len(nums)+1):
if i not in nums:
return i
``````

• @nitinsurya finding a number in a set is O(n) operation in worst case, only in average case it is O(1) depending on the input size and hash table size which implements this set.

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