# First Missing Positive

Key takeaway: One of the best searching problems to learning step-by-step optimization using various approaches. In-place hashing solution is an intuitive problem-solving approach that uses the same input array to process values and generate correct output.

### Let’s understand the problem

Given an array that includes positive and negative numbers, write a program to find the smallest missing positive integer.

Example 1

Input: X[] = [2, -9, 5, 11, 1, -10, 7], Output: 3

Explanation: Consecutive positive integers 1 and 2 are present in the array. So the first missing positive is 3.

Example 2

Input: X[] = [2, 3, -2, 1], Output: 4

Explanation: Consecutive positive integers 1, 2, and 3 are present in the array. So the first missing positive is 4.

Example 3

Input: X[] = [4, 6, 5, 3, 8], Output: 1

Explanation: All values are positive, but the first positive integer 1 is missing in the input. So the output is 1.

### Discussed solution approaches

• A brute force approach  using two nested loops
• Using sorting and single scan
•  Using a hash table
• Using in-place hashing
• Using the idea of partition and In-place Hashing

### A brute force approach  using two nested loops

Solution idea

The positive number starts from 1 on the number scale. If the first missing positive number is k, then all the positive numbers from 1 to k - 1 will be present in the array.

So the critical question is: What would be the maximum possible missing positive value in an array of size n? Think! The answer would be simple: in the n size array, the maximum possible missing positive number would be n + 1 i.e, this is the scenario when all numbers from 1 to n are present in the array.

So one basic approach would be to search all positive integers starting from 1 to n + 1 linearly in the array. If any positive integer is missing in the sequence, we will return that value as an output.

Solution pseudocode

``````int firstMissingPositive(int X[], int n)
{
for (int i = 1; i <= n + 1; i = i + 1)
{
bool missingFlag = true
for (int j = 0; j < n; j = j + 1)
{
if (X[j] == i)
{
missingFlag = false
break
}
}
if (missingFlag == true)
return i
}
}
``````

Solution analysis

We are running two nested loops doing constant operations at each iteration. Here both loops will run n times in the worst-case. So the time complexity in the worst-case = O(n)*O(n)*O(1) = O(n^2). We are using constant extra space, so space complexity = O(1).

### Using sorting and single scan

Solution idea and steps

If we sort the input array, then all the positive numbers get lined up in increasing order, and we can search linearly to find the first missing positive integer. For sorting, we can use efficient sorting algorithms like heap sort, merge sort, or quicksort.

1. We sort the array. Suppose we are using heap sort which is an in-place O(nlogn) sorting algorithm.
2. Now we run a loop and skip the negative numbers and zero present in the starting positions of the array. Starting from the first index, when we find X[i] < 1, we increment i by 1. By the end of this loop, the index i will be present at the first positive number.
3. Now we run another loop from j = i to n - 1 to find the first missing positive number. Outside the loop, we also declare the variable missingPositive to track to the first missing positive. The missingPositive will be initialized with the smallest positive number i.e. 1 at the start.
4. We start comparing the missingPositive with X[i]. If (missingPositive == X[i]), we move to the next positive number in the sequence by incrementing missingPositive by 1.
5. We continue the above process until we find some value X[i] > missingPositive. At this stage, the value stored in missingPositive would be the first missing positive. So we return missingPositive.

Solution pseudocode

``````int firstMissingPositive(int X[], int n)
{
heapSort(X, n)
int i = 0
While (X[i] < 1)
i = i + 1
missingPositive = 1
for (int j = i; j < n; j = j + 1)
{
if (missingPositive == X[i])
missingPositive = missingPositive + 1
else if (X[i] > missingPositive)
return missingPositive
}
return missingPositive
}
``````

Solution analysis

Time complexity = Time complexity of sorting + Linear scan for finding the first missing positive = O(nlogn) + O(n) = O(nlogn). Space complexity = O(1) (Think!).

### Using a hash table

Solution idea

Now the critical question is: can we improve the time complexity further? As empahsized in the brute force approach, this is a searching problem where we need to look for positive integers starting from 1 to n + 1. Can we think of optimizing the brute force approach? Think!

To enhance the searching efficiency, one idea would be to use a hash table instead of a linear search. The hash table can help us to perform searching and insertion efficiently in the O(1) average.

Solution steps

• We create a hash table H of size n.
• Now we run a loop from i = 0 to n - 1 and insert all elements in the hash table H.
• Now we another loop from i = 1 to n + 1 to search all the possible positive numbers from 1 to n + 1 in the hash table H. If any positive integer i is not present in H, we return i. Otherwise, we move to the next iteration.

Solution Pseudocode

``````int firstMissingPositive(int X[], int n)
{
HashTable H
for (int i = 0; i < n; i = i + 1)
H[X[i]] = true
for (int i = 1; i <= n + 1; i = i + 1)
{
if (H[i] == false)
return i
}
}
``````

Solution analysis

Time complexity = Inserting n elements into the hash table + Searching n elements into the hash table = O(n) + O(n) = O(n). Space complexity = O(n), for the hash table of size n.

### Using in-place hashing

Solution idea and steps

Now the critical question is: can we solve this problem in place? Let’s think!

• We need to identify the first missing number in the sequence of 1 to n + 1. For this purpose, we can use the indexes of the same array to mark the presence of the numbers in the sequence and put each number in its correct place, i.e X [X[i] - 1] == X[i]. For example, we can place 1 at X, 2 at X, 3 at A, and so on.
• For placing the numbers in their right place, we scan the array X[] from the start and check if the number X[i] is in the range of 1 to n and already present at index X[i] - 1 or not. If the first condition is true and the second is false, we swap X[i] with the number present at the index X[i] - 1. Otherwise, we ignore the current number and move to the next index.
• Now we need to scan the modified array again to find the first index where X[i] != i + 1. For some value of i, if X[i] != i + 1, then we return i + 1 as the first missing positive. If we didn't find any such index, all numbers from 1 to n are present in the array, and we need to return the value n + 1 as the first missing positive.

The best part of this idea is: we are modifying the same array and storing the results to get the correct output.

Solution Pseudocode

``````int firstMissingPositive(int X[], int n)
{
int i = 0
while (i < n)
{
if (X[i] > 0 && X[i] <= n && X[X[i] - 1] != X[i])
swap(X[i], X[X[i] - 1])
else
i = i + 1
}

for (int i = 0; i < n; i = i + 1)
{
if (X[i] != i + 1)
return i + 1
}
return n + 1
}
``````

Solution analysis

We are running two separate single loops n times. So time complexity in the worst case = Time complexity of modifying the array + Time complexity of searching first missing positive = O(n) + O(n) = O(n)

Space complexity = O(1), as we use the same input array to place the elements at their correct position. That's why such approaches are always premium!

### Using partition process and in-place hashing

Solution idea and steps

This idea is similar to the above approach, but there is a difference! We first divide the array into two parts: the first part consists of positive numbers, and the second part consists of negative numbers. We are assuming the array can be modified.

• We partition the input array into a group of positive and negative numbers using the partition algorithm similar to the quick sort. After the partition, the positive number starts from index 0 and ends at index positiveEnd - 1. Here positiveEnd value is returned by the partition process.
• Now we traverse the array from 0 to positiveEnd - 1. If X[i] > positiveEnd, then we do nothing. If not, we make the sign of the element at the index X[i] - 1 negative (Think!)
• Finally, we traverse the array once more from index 0 to positiveEnd -1. If we encounter a positive element at some index i, we output i + 1 as the first missing positive. If we do not encounter any positive element, then integers from 1 to positiveEnd are present in the array and we return positiveEnd + 1 as an output. Note: It can also be the case that all the numbers are non-positive, making positiveEnd = 0. The output positiveEnd + 1 = 1 remains correct.

We can better understand this approach via an example.

Initial Array: 1 -1 -5 -3 3 4 2 8

Step 1: Partition: 1 8 2 4 3 | -3 -5 -1, positiveEnd = 5

Step 2: Modifying the input array.

• 1 < positiveEnd, we change the sign of value at index 1 - 1 = 0

Modified array: -1 8 2 4 3 | -3 -5 -1

• 8 > positiveEnd, we do nothing and move to the next element.
• 2 < positiveEnd, we change the sign of value at index 2 - 1 = 1

Modified array: -1 -8 2 4 3 | -3 -5 -1

• 4 < positiveEnd, we change the sign of value at index 4 - 1 = 3

Modified array: -1 -8 2 -4 3 | -3 -5 -1

• 3 < positiveEnd, we change the sign of value at index 3 - 1 = 2

Modified array: -1 -8 -2 -4 3 | -3 -5 -1

Step 3: Traversing from index 0 to positiveEnd, we find X = 3 > 0. The answer is 4 + 1 = 5

Why this approach is working perfectly? What is the critical reason behind it? We are leaving this intuition for the learners to think! Enjoy thinking!

Solution Pseudocode

``````int firstMissingPositive(int X[], int n)
{
int positiveEnd = partition(X, n)
for (int i = 0; i < positiveEnd; i = i + 1)
{
if (X[i] - 1 < positiveEnd)
X[X[i] - 1] = - X[X[i] - 1]
}

for (int i = 0; i < positiveEnd; i = i + 1)
{
if (X[i] > 0)
return i + 1
}
return positiveEnd + 1
}

int partition(int X[], int n)
{
int j = n - 1, i = 0
while (i <= j)
{
if (A[i] <= 0)
{
swap(A[i], A[j])
j = j - 1
}
else
i = i + 1
}
return i
}
``````

Solution analysis

Time complexity = Time complexity of the partition + time complexity of modifying the input array + Linear scan to find the missing positive = O(n) + O(n) + O(n) = O(n)

Space complexity = O(1), we are solving the problem in place using the same input array!

### Critical ideas to think!

• In the 4th and 5th approach, why are we not storing the occurrences of numbers greater than n?
• What if we need to find the smallest missing number?
• What if we need to find the largest missing negative number?
• Is it possible to output all the numbers missing in the array in the range [1, n]?
• In the 2nd approach, our array is sorted. Can we think to optimize the search by applying the binary search algorithm?
• Can we solve the problem using some other approach?
• Will the 4th and 5th approaches work if the input array contains large numbers?
• In the 3rd approach, do we need to store all non-positive integers?
• What if all the elements that are present in the array are negative? Do the above algorithms cover this condition?

### Comparisons of time and space complexities

• Using two nested loops: Time = O(n^2), Space = O(1)
• Using sorting and single scan: Time = O(nlogn), Space = O(1)
•  Using a Hash Table: Time = O(n), Space = O(n)
• Using idea similar to counting sort: Time = O(n), Space = O(1)
• Using partition and In-place Hashing: Time = O(n), Space = O(1)

### Suggested coding problems to practice

• The kth missing element in an unsorted array.
• Smallest prime number missing in an array.
• Find the missing integer in an array if the mean is given.
• Find the missing number in another array which is a shuffled copy.

Enjoy learning, Enjoy algorithms!