Count numbers which can be constructed using two numbers

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.
Examples :

Input  : n = 10, x = 2, y = 3
Output : 9
We can form 9 out of 10 numbers using 2 and 3
2 = 2, 3 = 3, 4 = 2+2, 5 = 2+3, 6 = 3+3
7 = 2+2+3, 8 = 3+3+2, 9 = 3+3+3
and 10 = 3+3+2+2. 
 
Input  : n = 10, x = 5, y = 7
Output : 3
We can form 3 out of 10 numbers using 5 and 7
The numbers are 5, 7 and 10
Input  : n = 15, x = 5, y = 7
Output : 6
We can form 6 out of 10 numbers using 5 and 7.
The numbers are 5, 7, 10, 12, 14 and 15.
Input  : n = 15, x = 2, y = 4
Output : 7

A 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.

 Below is the implementation.

C++

// C++ program to count all numbers that can
// be formed using two number numbers x an y
#include<bits/stdc++.h>
using namespace std;

// Returns count of numbers from 1 to n that can be formed
// using x and y.
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;
}

// Driver code
int main()
{
    int n = 15, x = 5, y = 7;
    cout << countNums(n, x, y);
    return 0;
}

Java

// Java program to count all numbers that can
// be formed using two number numbers x an y

class gfg{
// Returns count of numbers from 1 to n that can be formed
// using x and y.
static 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
    boolean[] arr=new boolean[n+1];

    // 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=Math.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;
}

// Driver code
public static void main(String[] args)
{
    int n = 15, x = 5, y = 7;
    System.out.println(countNums(n, x, y));
}
}
// This code is contributed by mits

Python3

# Python3 program to count all numbers
# that can be formed using two number
# numbers x an y

# Returns count of numbers from 1
# to n that can be formed using x and y.
def countNums(n, x, 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
    arr = [False for i in range(n + 2)]

    # x and y can be formed using x and y.
    if(x <= n):
        arr[x] = True
    if(y <= n):
        arr[y] = True

    # Initialize result
    result = 0

    # Traverse all numbers and increment
    # result if a number can be formed
    # using x and y.
    for i in range(min(x, y), n + 1):

        # 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 = result + 1

    return result

# Driver code
n = 15
x = 5
y = 7
print(countNums(n, x, y))

# This code is contributed by
# Sanjit_Prasad

C#

// C# program to count all numbers that can
// be formed using two number numbers x an y

using System;

public class GFG{
    // Returns count of numbers from 1 to n that can be formed
// using x and y.
static 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
    bool []arr=new bool[n+1];

    // 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=Math.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;
}

// Driver code
    static public void Main (){
        int n = 15, x = 5, y = 7;
        Console.WriteLine(countNums(n, x, y));
    }
}

Javascript

<script>
// javascript program to count all numbers that can
// be formed using two number numbers x an y

// Returns count of numbers from 1 to n that can be formed
// using x and y.
function countNums(n, x, 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
    arr= Array(n+1).fill(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
    var result = 0;

    // Traverse all numbers and increment
    // result if a number can be formed using
    // x and y.
    for (i = Math.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;
}

// Driver code
var n = 15, x = 5, y = 7;
document.write(countNums(n, x, y));

// This code is contributed by Princi Singh
</script>

PHP

<?php
// PHP program to count all numbers
// that can be formed using two
// number numbers x an y

// Returns count of numbers from
// 1 to n that can be formed
// using x and y.
function countNums($n, $x, $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
    $arr = array_fill(0, $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
    $result = 0;

    // Traverse all numbers and increment
    // result if a number can be formed
    // using x and y.
    for ($i = min($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;
}

// Driver code
$n = 15;
$x = 5;
$y = 7;
echo countNums($n, $x, $y);

// This code is contributed by mits
?>

Output

6

Time Complexity: O(n) 
Auxiliary Space: O(n)

Approach 2: Using DP:

This code implements a dynamic programming approach to solve a problem of counting the number of integers that can be formed using two given integers x and y, up to a given integer n.

• The function countNumsDP takes four arguments: n, x, y, and a reference to a vector called dp. The integer n is the upper limit up to which the numbers need to be counted, and x and y are the two given integers that can be used to form other integers. The vector dp is used to store the results of subproblems, which can be reused in later computations to avoid redundant computations.
  • The function first checks if the base case n=0 is reached, and returns 0. If the result for n has already been computed and stored in dp, then it simply returns the result without performing further computations.
  • The function then creates a boolean array called arr, where arr[i] represents whether the number i can be formed using x and y or not. It sets arr[x] and arr[y] to true if x and y are less than or equal to n.
  • Next, the function iterates over all numbers from min(x,y) to n, and if the number i can be formed using x and y, it checks whether i+x and i+y can also be formed using x and y. If they can be formed, the function increments the result counter and sets arr[i+x] and arr[i+y] to true.
  • Finally, it stores the result in dp[n] and returns the result.

This approach utilizes the previously computed results stored in the dp table to avoid redundant computations, thus improving the efficiency of the algorithm.

C++

#include<bits/stdc++.h>
using namespace std;

int countNumsDP(int n, int x, int y, vector<int>& dp) {
    if(n == 0) return 0; // Base case
    if(dp[n] != -1) return dp[n]; // If result already computed, return it

    vector<bool> arr(n+1, false); // Array to store intermediate results
    if(x <= n) arr[x] = true; // x and y can be formed using x and y.
    if(y <= n) arr[y] = true;

    int result = 0; // Initialize result

    // 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(arr[i]) { // If i can be formed using x and y
            if(i+x <= n) arr[i+x] = true; // Then i+x and i+y can also be formed using x and y.
            if(i+y <= n) arr[i+y] = true;

            result++; // Increment result
        }
    }

    dp[n] = result; // Store the result in dp table
    return result;
}

int main() {
    int n = 15, x = 5, y = 7;
    vector<int> dp(n+1, -1); // Initialize dp table with -1
    cout << countNumsDP(n, x, y, dp) << endl; // Call DP function
    return 0;
}

Java

import java.util.Arrays;

public class GFG {

    // Function to count the number of integers that can be formed using x and y
    static int countNumsDP(int n, int x, int y, int[] dp) {
        if (n == 0)
            return 0; // Base case
        if (dp[n] != -1)
            return dp[n]; // If result already computed, return it

        boolean[] arr = new boolean[n + 1]; // Array to store intermediate results
        if (x <= n)
            arr[x] = true; // x and y can be formed using x and y.
        if (y <= n)
            arr[y] = true;

        int result = 0; // Initialize result

        // Traverse all numbers and increment result if a number can be formed using x and y.
        for (int i = Math.min(x, y); i <= n; i++) {
            if (arr[i]) { // If i can be formed using x and y
                if (i + x <= n)
                    arr[i + x] = true; // Then i+x and i+y can also be formed using x and y.
                if (i + y <= n)
                    arr[i + y] = true;

                result++; // Increment result
            }
        }

        dp[n] = result; // Store the result in dp table
        return result;
    }

    public static void main(String[] args) {
        int n = 15, x = 5, y = 7;
        int[] dp = new int[n + 1];
        Arrays.fill(dp, -1); // Initialize dp table with -1
        System.out.println(countNumsDP(n, x, y, dp)); // Call DP function
    }
}

Python3

def countNumsDP(n, x, y, dp):
    if n == 0:
        return 0  # Base case
    if dp[n] != -1:
        return dp[n]  # If result already computed, return it

    arr = [False] * (n + 1)  # Array to store intermediate results
    if x <= n:
        arr[x] = True  # x and y can be formed using x and y.
    if y <= n:
        arr[y] = True

    result = 0  # Initialize result

    # Traverse all numbers and increment result if a number can be formed using x and y.
    for i in range(min(x, y), n + 1):
        if arr[i]:  # If i can be formed using x and y
            if i + x <= n:
                # Then i+x and i+y can also be formed using x and y.
                arr[i + x] = True
            if i + y <= n:
                arr[i + y] = True

            result += 1  # Increment result

    dp[n] = result  # Store the result in dp table
    return result


def main():
    n = 15
    x = 5
    y = 7
    dp = [-1] * (n + 1)  # Initialize dp table with -1
    print(countNumsDP(n, x, y, dp))  # Call DP function


if __name__ == "__main__":
    main()

# This code is contributed by shivamgupta310570

C#

using System;
using System.Collections.Generic;

class GFG {
    static int CountNumsDP(int n, int x, int y, List<int> dp) {
        if (n == 0) return 0; // Base case
        if (dp[n] != -1) return dp[n]; // If result already computed, return it

        List<bool> arr = new List<bool>(n + 1);
        for (int i = 0; i <= n; i++) {
            arr.Add(false);
        }

        if (x <= n) arr[x] = true;
        if (y <= n) arr[y] = true;

        int result = 0;

        for (int i = Math.Min(x, y); i <= n; i++) {
            if (arr[i]) {
                if (i + x <= n) arr[i + x] = true;
                if (i + y <= n) arr[i + y] = true;

                result++;
            }
        }

        dp[n] = result;
        return result;
    }

    public static void Main(string[] args) {
        int n = 15, x = 5, y = 7;
        List<int> dp = new List<int>(n + 1);
        for (int i = 0; i <= n; i++) {
            dp.Add(-1);
        }

        Console.WriteLine(CountNumsDP(n, x, y, dp));
    }
}

Javascript

//Javascript code for the above approach
function countNumsDP(n, x, y, dp) {
    if (n === 0) return 0; // Base case
    if (dp[n] !== -1) return dp[n]; // If result already computed, return it

    const arr = new Array(n + 1).fill(false); // Array to store intermediate results
    if (x <= n) arr[x] = true; // x and y can be formed using x and y.
    if (y <= n) arr[y] = true;

    let result = 0; // Initialize result

    // Traverse all numbers and increment result if a number can be formed using x and y.
    for (let i = Math.min(x, y); i <= n; i++) {
        if (arr[i]) { // If i can be formed using x and y
            if (i + x <= n) arr[i + x] = true; // Then i+x and i+y can also be
                                               // formed using x and y.
            if (i + y <= n) arr[i + y] = true;

            result++; // Increment result
        }
    }

    dp[n] = result; // Store the result in dp table
    return result;
}

function main() {
    const n = 15, x = 5, y = 7;
    const dp = new Array(n + 1).fill(-1); // Initialize dp table with -1
    console.log(countNumsDP(n, x, y, dp)); // Call DP function
}

main();

Output

6

Time complexity: O(n * min(x, y))
Auxiliary Space: O(n)