07 Union of two Sorted Arrays

Union of two arrays | Practice | GeeksforGeeks

GeeksFor Geeks QUesiton

Using Set

int doUnion(int a[], int n, int b[], int m)  {
    set<int> s(a, a + n);
    for(int i = 0; i < m; i++)
        s.insert(b[i]);
        
    return s.size();
}

Sorting takes $O(n \log mn)$

Inserting n elements in set takes

$\log(1) + \log(2) + ... + \log(n)$ time

$=\log(1\times2\times ... \times n)$$=\log(n!)$

So, for inserting n and m elements it takes

$\log(n!\times m!)$ time

So, Time Complexity = $O(\log(n!\times m!))$

Space Complexity = $O(m + n)$​

Using Look Up Table

int doUnion(int a[], int n, int b[], int m)  {
    // find the largest element from both the arrays
    int maxElement = INT_MIN;
    for(int i = 0; i < n; i++)
        maxElement = max(maxElement, a[i]);
    for(int i = 0; i < m; i++)
        maxElement = max(maxElement, b[i]);
        
    // create a lookup table as big as the largest element
    vector<bool> present(maxElement + 1, false);
    
    // result
    int count = 0;
    
    // for every element in first array
    for(int i = 0; i < n; i++)
        // if it has not been encountered yet
        if(!present[a[i]]) {
            // mark it as seen
            present[a[i]] = true;
            // count it once
            count++;
        } // any subsequent encounters will not be counted
        
    // for every element in second array
    for(int i = 0; i < m; i++)
        // if it has not been encoutered yet.
        // This includes encounters from 1st array
        if(!present[b[i]]) {
            // mark it as seen
            present[b[i]] = true;
            // count it once.
            count++;
        } // any subsequent encouters will not be counted
        
    // final result in count
    return count;
}

Time Complexity: $O(m + n)$

Space complexity: $O(max(a, b))$