Count numbers which can be constructed using two numbers - GeeksforGeeks


Count numbers which can be constructed using two numbers - GeeksforGeeks
Given three positive integers x, y and n, the task is to find count of all numbers from 1 to n that can be formed using x and y. A number can be formed using x and y if we can get it by adding any number of occurrences of x and/or y.


simple solution is to write a recursive code that starts with 0 and makes two recursive calls. One recursive call adds x and other adds y. This way we count total numbers. We need to make sure a number is counted multiple times.
An efficient solution solution is to use a boolean array arr[] of size n+1. An entry arr[i] = true is going to mean that i can be formed using x and y. We initialize arr[x] and arr[y] as true if x and y are smaller than or equal to n. We start traversing the array from smaller of two numbers and mark all numbers one by one that can be formed using x and y.
Time Complexity : O(n)
Auxiliary Space : O(n)

int countNums(int n, int x, int y)
{
    // Create an auxiliary array and initialize it
    // as false. An entry arr[i] = true is going to
    // mean that i can be formed using x and y
    vector<bool> arr(n+1, false);
 
    // x and y can be formed using x and y.
    if (x <= n)
        arr[x] = true;
    if (y <= n)
        arr[y] = true;
 
    // Initialize result
    int result = 0;
 
    // Traverse all numbers and increment
    // result if a number can be formed using
    // x and y.
    for (int i=min(x, y); i<=n; i++)
    {
        // If i can be formed using x and y
        if (arr[i])
        {
            // Then i+x and i+y can also be formed
            // using x and y.              
            if (i+x <= n)
                arr[i+x] = true;
            if (i+y <= n)
                arr[i+y] = true;
 
            // Increment result
            result++;
        }
    }
    return result;
}
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