First Missing Non Negative Integer

Last updated: 17th Jan, 2021

Problem Statment

Given an unsorted integer array, find the first missing non negative integer.

Example 1:

[1,2,0]

>> 3

Approach 1:

The bruteforce approach for such a problem involves, sorting the given array.
After sorting, iterate over the array of length 'n'. For each i:0->n-1, if array[i]!=i, then the current value of i, is the first missing non negative integer and can be returned. If the array is iterated and for all valid i(0->n-1), array[i]=i, the first missing non negative integer is n.

Time and Space Complexity :
Time - O(nlogn)
Space - O(1)

C++ Code

void missing_integer(vector<int>a)
{
	sort(a.begin(),a.end());
	for(int i=0;i<a.size();i++){
    if(a[i]!=i){
      cout<<i<<'\n';
      return;
      }
    }
	cout<< a.size()<<'\n';
}

Approach 2:

This solution can be further optimised.
The basic idea is for any n non negative numbers, the first missing non negative number must be within [0,n]. The reasoning is similar to putting k balls into k+1 bins, there must be a bin empty, the empty bin can be viewed as the missing number.

Time and Space Complexity :
Time - O(n)
Space - O(1)

C++ Code

void missing_integer(vector<int>a)
{
  int n = a.size();
  for(int i=0;i<n;i++){
    while(a[i]>=0 && a[i]<n && a[a[i]]!=a[i]){ 
      //Elements in array can vary from 0 to n-1
      //An array of size n has indixes varying from 0 to n-1 too
      //Place element with value i, at index i
      swap(a[i], a[a[i]]);
      }
  }
  for(int i=0;i<n;i++){
    if(a[i]!=i){
      cout<<i<<'\n';
      return;
    }
  }
  cout<< a.size()<<'\n';
}