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Print all Good numbers in given range

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Given a digit ‘d’ and a range [L, R] where L < R, print all good numbers in given range that don’t contain digit ‘d’. A number is good if its every digit is larger than the sum of digits which are on the right side of that digit. For example 9620 is good number because 2 > 0, 6 > 2+0 and 9 > 6+2+0.

Example:  

Input:  L = 410, R = 520, d = 3
Output: 410 420 421 510 520 
All the numbers in output are good (every digit is more
than sum of digits on right of it) and don't have digit 3.

Input:  L = 410, R = 520, d = 1
Output: 420 430 520 
All the numbers in output are good (every digit is more
than sum of digits on right of it) and don't have digit 1.
Recommended Practice

The idea is to traverse all numbers in the given range. For every number, traverse all digits. While traversing keep track of digit sum so far. At any point, if the previous sum becomes more than or equal to sum, return false. Also, if current digit becomes ‘d’, return false.

Below is the implementation of the idea. 

C++




// C++ program to print good numbers in a given range [L, R]
#include<bits/stdc++.h>
using namespace std;
 
// To check whether n is a good number and doesn't contain
// digit 'd'.
bool isValid(int n, int d)
{
    // Get last digit and initialize sum from right side
    int digit = n%10;
    int sum = digit;
 
    // If last digit is d, return
    if (digit == d)
    return false;
 
    // Traverse remaining digits
    n /= 10;
    while (n)
    {
        // Current digit
        digit = n%10;
 
        // If digit is d or digit is less than or
        // equal to sum of digits on right side
        if (digit == d || digit <= sum)
            return false;
 
        // Update sum and n
        else
        {
            sum += digit;
            n /= 10;
        }
    }
    return 1;
}
 
// Print Good numbers in range [L, R]
void printGoodNumbers(int L, int R, int d)
{
// Traverse all numbers in given range
for (int i=L; i<=R; i++)
{
    // If current numbers is good, print it.
    if (isValid(i, d))
        cout << i << " ";
}
}
 
// Driver program
int main()
{
    int L = 410, R = 520, d = 3;
 
    // Print good numbers in [L, R]
    printGoodNumbers(L, R, d);
 
    return 0;
}


Java




// Java program to print good numbers in a given range [L, R]
import java.io.*;
 
class Numbers
{
    // Function to check whether n is a good number and doesn't contain
    // digit 'd'
    static boolean isValid(int n, int d)
    {
        // Get last digit and initialize sum from right side
        int digit = n%10;
        int sum = digit;
 
        // If last digit is d, return
        if (digit == d)
        return false;
 
        // Traverse remaining digits
        n /= 10;
        while (n>0)
        {
            // Current digit
            digit = n%10;
     
            // If digit is d or digit is less than or
            // equal to sum of digits on right side
            if (digit == d || digit <= sum)
                return false;
 
            // Update sum and n
                else
                {
                    sum += digit;
                    n /= 10;
                }
        }
    return true;
    }
     
    // Print Good numbers in range [L, R]
    static void printGoodNumber(int L, int R, int d)
    {
        // Traverse all numbers in given range
        for(int i=L;i<=R;i++)
        {
            // If current numbers is good, print it
            if(isValid(i, d))
                System.out.print(i+" ");
        }
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int L = 410, R = 520, d = 3;
         
        // Print good numbers in [L, R]
        printGoodNumber(L, R, d);
    }
}


Python3




# Python3 program to print good
# numbers in a given range [L, R]
 
# Function to check whether n is
# a good number and doesn't contain
# digit 'd'
def isValid(n, d):
     
    # Get last digit and initialize
    # sum from right side
    digit = n % 10;
    sum = digit;
 
    # If last digit is d, return
    if (digit == d):
        return False;
 
    # Traverse remaining digits
    n = int(n / 10);
    while (n > 0):
         
        # Current digit
        digit = n % 10;
     
        # If digit is d or digit is
        # less than or equal to sum
        # of digits on right side
        if(digit == d or digit <= sum):
            return False;
             
        # Update sum and n
        else:
            sum += digit;
            n = int(n / 10);
    return True;
     
# Print Good numbers in range [L, R]
def printGoodNumber(L, R, d):
     
    # Traverse all numbers
    # in given range
    for i in range(L, R + 1):
         
        # If current numbers is good,
        # print it
        if(isValid(i, d)):
            print(i, end = " ");
     
# Driver Code
L = 410;
R = 520;
d = 3;
         
# Print good numbers in [L, R]
printGoodNumber(L, R, d);
 
# This code is contributed by mits


C#




// C# program to print good numbers in a
// given range [L, R]
using System;
 
class GFG {
     
    // Function to check whether n is a good
    // number and doesn't contain digit 'd'
    static bool isValid(int n, int d)
    {
         
        // Get last digit and initialize
        // sum from right side
        int digit = n % 10;
        int sum = digit;
 
        // If last digit is d, return
        if (digit == d)
            return false;
 
        // Traverse remaining digits
        n /= 10;
        while (n > 0)
        {
             
            // Current digit
            digit = n % 10;
     
            // If digit is d or digit is
            // less than or equal to sum of
            // digits on right side
            if (digit == d || digit <= sum)
                return false;
 
            // Update sum and n
            else
            {
                sum += digit;
                n /= 10;
            }
        }
         
    return true;
    }
     
    // Print Good numbers in range [L, R]
    static void printGoodNumber(int L,
                               int R, int d)
    {
         
        // Traverse all numbers in given range
        for(int i = L; i <= R; i++)
        {
             
            // If current numbers is good,
            // print it
            if(isValid(i, d))
                Console.Write(i+" ");
        }
    }
     
    // Driver program
    public static void Main ()
    {
        int L = 410, R = 520, d = 3;
         
        // Print good numbers in [L, R]
        printGoodNumber(L, R, d);
    }
}
 
//This code is contributed by vt_m.


PHP




<?php
// PHP program to print good
// numbers in a given range [L, R]
 
// To check whether n is a good
// number and doesn't contain digit 'd'.
function isValid($n, $d)
{
    // Get last digit and initialize
    // sum from right side
    $digit = $n % 10;
    $sum = $digit;
 
    // If last digit is d, return
    if ($digit == $d)
    return false;
 
    // Traverse remaining digits
    $n = (int)($n / 10);
    while ($n)
    {
        // Current digit
        $digit = $n % 10;
 
        // If digit is d or digit is less
        // than or equal to sum of digits
        // on right side
        if ($digit == $d || $digit <= $sum)
            return false;
 
        // Update sum and n
        else
        {
            $sum += $digit;
            $n = (int)($n / 10);
        }
    }
    return 1;
}
 
// Print Good numbers in range [L, R]
function printGoodNumbers($L, $R, $d)
{
// Traverse all numbers in given range
for ($i = $L; $i <= $R; $i++)
{
    // If current numbers is good,
    // print it.
    if (isValid($i, $d))
        echo $i . " ";
}
}
 
// Driver Code
$L = 410;
$R = 520;
$d = 3;
 
// Print good numbers in [L, R]
printGoodNumbers($L, $R, $d);
 
// This code is contributed by mits
?>


Javascript




<script>
 
// Javascript program to print good numbers
// in a given range [L, R]class Numbers
 
// Function to check whether n is a good
// number and doesn't contain digit 'd'
function isValid(n, d)
{
     
    // Get last digit and initialize
    // sum from right side
    var digit = n % 10;
    var sum = digit;
 
    // If last digit is d, return
    if (digit == d)
        return false;
 
    // Traverse remaining digits
    n = parseInt(n / 10);
    while (n > 0)
    {
         
        // Current digit
        digit = n%10;
 
        // If digit is d or digit is less than or
        // equal to sum of digits on right side
        if (digit == d || digit <= sum)
            return false;
             
        // Update sum and n
        else
        {
            sum += digit;
            n = parseInt(n / 10);
        }
    }
    return true;
}
 
// Print Good numbers in range [L, R]
function printGoodNumber(L, R, d)
{
     
    // Traverse all numbers in given range
    for(i = L; i <= R; i++)
    {
         
        // If current numbers is good, print it
        if (isValid(i, d))
            document.write(i + " ");
    }
}
 
// Driver code
var L = 410, R = 520, d = 3;
 
// Print good numbers in [L, R]
printGoodNumber(L, R, d);
 
// This code is contributed by shikhasingrajput
 
</script>


Output

410 420 421 510 520 

Time Complexity: O((l-r) * log10 n) , takes O(log10 n) time to check if a number is a good number
Auxiliary Space: O(1) 



Last Updated : 09 Feb, 2023
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