Is below sorting algorithm O(n)?

Question

Algorithm:

1. insert element counts in a map
2. start from first element
3. if first is present in a map, insert in output array (total number of count), increment first
4. if first is not in a map, find next number which is present in a map

Complexity: O(max element in array) which is linear, so, O(n).

vector<int> sort(vector<int>& can) {
    unordered_map<int,int> mp;
    int first = INT_MAX;
    int last = INT_MIN;
    for(auto &n : can) {
        first = min(first, n);
        last = max(last, n);
        mp[n]++;
    }
    vector<int> out;
    while(first <= last) {
        while(mp.find(first) == mp.end()) first ++;
        int cnt = mp[first];
        while(cnt--) out.push_back(first);
        first++;
    }
    return out;
}

Answer 1

Complexity: O(max element in array) which is linear, so, O(n).

No, it's not O(n). The while loop iterates last - first + 1 times, and this quantity depends on the array's contents , not the array's length .

Usually we use n to mean the length of the array that the algorithm works on. To describe the range (ie the difference between the largest and smallest values in the array), we could introduce a different variable r, and then the time complexity is O(n + r), because the first loop populating the map iterates O(n) times, the second loop populating the vector iterates O(r) times, and its inner loop which counts down from cnt iterates O(n) times in total.

Another more formal way to define n is the "size of the input", typically measured in the number of bits that it takes to encode the algorithm's input. Suppose the input is an array of length 2, containing just the numbers 0 and M for some number M. In this case, if the number of bits used to encode the input is n, then the number M can be on the order of O(2 ⁿ ), and the second loop does that many iterations; so by this formal definition the time complexity is exponential.