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

• Last Updated : 08 Nov, 2021

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

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 using namespace std; // Function to return the sum of the digits of nint 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 nint 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 codeint 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 nstatic 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 nstatic 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 ndef 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 ndef 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 codeif __name__ == "__main__" :         n = 824;    print(findNumber(n)) # This code is contributed by Ryuga

C#

 // C# implementation of the approachusing System; class GFG{// Function to return the sum// of the digits of nstatic 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 nstatic 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 codestatic public void Main (){    int n = 824;    Console.WriteLine (findNumber(n));}} // This code is contributed by @Tushil

PHP

 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 nfunction 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


Output:
819

Time Complexity: O(N)

Auxiliary Space: O(1)

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