Largest number less than N with digit sum greater than the digit sum of N

Last Updated : 09 May, 2023

Given an integer N, the task is to find the greatest number less than N such that the sum of its digits is greater than the sum of the digits of N. If the condition isn't satisfied for any number then print -1.
Examples:

Input: N = 100
Output: 99
99 is the largest number less than 100 sum of whose digits is greater than the sum of the digits of 100
Input: N = 49
Output: -1

Brute Force Approach:

To solve this problem we can start iterating from N-1 down to 1, and for each number, compute the sum of its digits using a loop. Then, we can compare this sum to the sum of the digits of N, and if it is greater, we have found the answer and we can return the current number. If the loop completes without finding a suitable number, we can return -1.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
int findNumber(int n)
{
    for (int i = n - 1; i > 0; i--) {
        int sum_i = 0, sum_n = 0;
        int temp_i = i, temp_n = n;
        while (temp_i > 0) {
            sum_i += temp_i % 10;
            temp_i /= 10;
        }
        while (temp_n > 0) {
            sum_n += temp_n % 10;
            temp_n /= 10;
        }
        if (sum_i > sum_n) {
            return i;
        }
    }
    return -1;
}

// Driver code
int main()
{
    int n = 824;
    cout << findNumber(n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class Main {
    // Function to return the greatest
    // number less than n such that
    // the sum of its digits is greater
    // than the sum of the digits of n
    static int findNumber(int n)
    {
        for (int i = n - 1; i > 0; i--) {
            int sum_i = 0, sum_n = 0;
            int temp_i = i, temp_n = n;
            while (temp_i > 0) {
                sum_i += temp_i % 10;
                temp_i /= 10;
            }
            while (temp_n > 0) {
                sum_n += temp_n % 10;
                temp_n /= 10;
            }
            if (sum_i > sum_n) {
                return i;
            }
        }
        return -1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 824;
        System.out.println(findNumber(n));
    }
}

Python3

# Function to return the greatest
# number less than n such that
# the sum of its digits is greater
# than the sum of the digits of n


def findNumber(n):
    for i in range(n-1, 0, -1):
        sum_i = 0
        sum_n = 0
        temp_i = i
        temp_n = n
        while temp_i > 0:
            sum_i += temp_i % 10
            temp_i //= 10
        while temp_n > 0:
            sum_n += temp_n % 10
            temp_n //= 10
        if sum_i > sum_n:
            return i
    return -1


# Driver code
n = 824
print(findNumber(n))

using System;

public class Program 
{

  // Function to return the greatest
  // number less than n such that
  // the sum of its digits is greater
  // than the sum of the digits of n
  public static int FindNumber(int n)
  {
    for (int i = n - 1; i > 0; i--) {
      int sum_i = 0, sum_n = 0;
      int temp_i = i, temp_n = n;
      while (temp_i > 0) {
        sum_i += temp_i % 10;
        temp_i /= 10;
      }
      while (temp_n > 0) {
        sum_n += temp_n % 10;
        temp_n /= 10;
      }
      if (sum_i > sum_n) {
        return i;
      }
    }
    return -1;
  }

  // Driver code
  public static void Main()
  {
    int n = 824;
    Console.WriteLine(FindNumber(n));
  }
}

JavaScript

function FindNumber(n) {
  for (let i = n - 1; i > 0; i--) {
    let sum_i = 0, sum_n = 0;
    let temp_i = i, temp_n = n;
    while (temp_i > 0) {
      sum_i += temp_i % 10;
      temp_i = Math.floor(temp_i / 10);
    }
    while (temp_n > 0) {
      sum_n += temp_n % 10;
      temp_n = Math.floor(temp_n / 10);
    }
    if (sum_i > sum_n) {
      return i;
    }
  }
  return -1;
}

function Main() {
  let n = 824;
  console.log(FindNumber(n));
}

Main();

Output

Time Complexity: O(N^2)
Auxiliary Space: O(1)

Approach: Start a loop from N-1 to 1 and check whether the sum of the digits of any number is greater than the sum of the digits of N. The first number that satisfies the condition is the required number.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the sum of the digits of n
int sumOfDigits(int n)
{
    int res = 0;

    // Loop for each digit of the number
    while (n > 0) {
        res += n % 10;
        n /= 10;
    }

    return res;
}

// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
int findNumber(int n)
{

    // Starting from n-1
    int i = n - 1;

    // Check until 1
    while (i > 0) {

        // If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n))
            return i;
        i--;
    }

    // If the condition is not satisfied
    return -1;
}

// Driver code
int main()
{
    int n = 824;
    cout << findNumber(n);

    return 0;
}

Java

//Java implementation of the approach

import java.io.*;

class GFG {
    // Function to return the sum of the digits of n
static int sumOfDigits(int n)
{
    int res = 0;

    // Loop for each digit of the number
    while (n > 0) {
        res += n % 10;
        n /= 10;
    }

    return res;
}

// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
static int findNumber(int n)
{

    // Starting from n-1
    int i = n - 1;

    // Check until 1
    while (i > 0) {

        // If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n))
            return i;
        i--;
    }

    // If the condition is not satisfied
    return -1;
}

// Driver code
    public static void main (String[] args) {

    int n = 824;
    System.out.println (findNumber(n));
    }
//This code is contributed by akt_mit    
}

Python3

# Python3 implementation of the approach

# Function to return the sum 
# of the digits of n
def sumOfDigits(n) :

    res = 0;

    # Loop for each digit of the number
    while (n > 0) :
        res += n % 10
        n /= 10

    return res;

# Function to return the greatest
# number less than n such that
# the sum of its digits is greater
# than the sum of the digits of n
def findNumber(n) :

    # Starting from n-1
    i = n - 1;

    # Check until 1
    while (i > 0) :

        # If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n)) :
            return i
            
        i -= 1

    # If the condition is not satisfied
    return -1;

# Driver code
if __name__ == "__main__" :
    
    n = 824;
    print(findNumber(n))

# This code is contributed by Ryuga

// C# implementation of the approach
using System;

class GFG
{
// Function to return the sum 
// of the digits of n
static int sumOfDigits(int n)
{
    int res = 0;

    // Loop for each digit of 
    // the number
    while (n > 0) 
    {
        res += n % 10;
        n /= 10;
    }

    return res;
}

// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
static int findNumber(int n)
{

    // Starting from n-1
    int i = n - 1;

    // Check until 1
    while (i > 0)
    {

        // If i satisfies the given condition
        if (sumOfDigits(i) > sumOfDigits(n))
            return i;
        i--;
    }

    // If the condition is
    // not satisfied
    return -1;
}

// Driver code
static public void Main ()
{
    int n = 824;
    Console.WriteLine (findNumber(n));
}
}

// This code is contributed by @Tushil

PHP

<?php
//PHP implementation of the approach

// Function to return the sum of 
// the digits of n
function sumOfDigits($n)
{
    $res = 0;

    // Loop for each digit of the number
    while ($n > 0) 
    {
        $res += $n % 10;
        $n /= 10;
    }

    return $res;
}

// Function to return the greatest
// number less than n such that
// the sum of its digits is greater
// than the sum of the digits of n
function findNumber($n)
{

    // Starting from n-1
    $i = $n - 1;

    // Check until 1
    while ($i > 0) 
    {

        // If i satisfies the given condition
        if (sumOfDigits($i) > sumOfDigits($n))
            return $i;
        $i--;
    }

    // If the condition is not satisfied
    return -1;
}

// Driver code
$n = 824;

echo findNumber($n);
    
// This code is contributed by Mukul singh
?>

JavaScript

<script>
// javascript implementation of the approach

    // Function to return the sum of the digits of n
    function sumOfDigits(n)
    {
        var res = 0;

        // Loop for each digit of the number
        while (n > 0)
        {
            res += n % 10;
            n = parseInt(n/10);
        }

        return res;
    }

    // Function to return the greatest
    // number less than n such that
    // the sum of its digits is greater
    // than the sum of the digits of n
    function findNumber(n)
    {

        // Starting from n-1
        var i = n - 1;

        // Check until 1
        while (i > 0)
        {

            // If i satisfies the given condition
            if (sumOfDigits(i) > sumOfDigits(n))
                return i;
            i--;
        }

        // If the condition is not satisfied
        return -1;
    }

    // Driver code
    var n = 824;
    document.write(findNumber(n));

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

Output

Time Complexity: O(N * log₁₀N)
Auxiliary Space: O(1)

Largest number not greater than N all the digits of which are odd

Samdare B

Improve

Article Tags :

Practice Tags :

Mathematical

Largest number less than N with digit sum greater than the digit sum of N

Similar Reads

Thank You!

What kind of Experience do you want to share?