Given a number n, find a number in range from 1 to n such that its sum is maximum. If there are several such integers, determine the **biggest** of them.

**Examples:**

Input: n = 100 Output: 99 99 is the largest number in range from 1 to 100 with maximum sum of digits. Input: n = 48 Output: 48 Explanation: There are two numbers with maximum digit sum. The numbers are 48 and 39 Since 48 > 39, it is our answer.

A **naive** approach is to iterate for all numbers from 1 to n and find out which number has maximum sum of digits. Time complexity of this solution is O(n).

An **efficient** approach is to iterate from n to 1. Do the following for each digit of current number, if the digit is not zero, reduce it by one and change all other digits to nine to the right of it. If the sum of digits in the resulting integer is strictly greater than the sum of the digits of the current answer, then update the answer with the resulting integer. If the sum of the resulting integer is same as the current answer, then if the resulting integer is more then current answer, update the current answer with the resulting integer.

**How to reduce a digit and change all other digits on its right to 9?**

Let x be our current number. We can find next number for current digit using below formula. In below formula, b is a power of 10 to represent position of current digit. After every iteration we reduce x to x/10 and change b to b * 10.

We use (x – 1) * b + (b – 1);

This line can further be explained as, if the number is x = 521 and b = 1, then

- (521 – 1) * 1 + (1-1) which gives you 520, which is the thing we need to do, reduce the position number by 1 and replace all other numbers to the right by 9.
- After x /= 10 gives you x as 52 and b*=10 gives you b as 10, which is again executed as (52-1)*(10) + 9 which gives you 519, which is what we have to do, reduce the current index by 1 and increase all other rights by 9.

## C++

// CPP program to find the // number with maximum digit // sum. #include <bits/stdc++.h> using namespace std; // function to calculate the // sum of digits of a number. int sumOfDigits(int a) { int sum = 0; while (a) { sum += a % 10; a /= 10; } return sum; } // Returns the maximum number // with maximum sum of digits. int findMax(int x) { // initializing b as 1 and // initial max sum to be of n int b = 1, ans = x; // iterates from right to // left in a digit while (x) { // while iterating this // is the number from // from right to left int cur = (x - 1) * b + (b - 1); // calls the function to // check if sum of cur is // more then of ans if (sumOfDigits(cur) > sumOfDigits(ans) || (sumOfDigits(cur) == sumOfDigits(ans) && cur > ans)) ans = cur; // reduces the number to one unit less x /= 10; b *= 10; } return ans; } // driver program int main() { int n = 521; cout << findMax(n); return 0; }

## Java

// Java program to find the // number with maximum digit // sum. import java.io.*; class GFG { // function to calculate the // sum of digits of a number. static int sumOfDigits(int a) { int sum = 0; while (a!=0) { sum += a % 10; a /= 10; } return sum; } // Returns the maximum number // with maximum sum of digits. static int findMax(int x) { // initializing b as 1 and // initial max sum to be of n int b = 1, ans = x; // iterates from right to // left in a digit while (x!=0) { // while iterating this // is the number from // from right to left int cur = (x - 1) * b + (b - 1); // calls the function to // check if sum of cur is // more then of ans if (sumOfDigits(cur) > sumOfDigits(ans) || (sumOfDigits(cur) == sumOfDigits(ans) && cur > ans)) ans = cur; // reduces the number to one unit less x /= 10; b *= 10; } return ans; } // driver program public static void main (String[] args) { int n = 521; System.out.println(findMax(n)); } } /*This article is contributed by Nikita Tiwari.*/

## Python3

# Python 3 program to # find the number # with maximum digit # sum. # function to calculate # the sum of digits of # a number. def sumOfDigits(a) : sm = 0 while (a!=0) : sm = sm + a % 10 a = a // 10 return sm # Returns the maximum number # with maximum sum of digits. def findMax(x) : # initializing b as 1 # and initial max sum # to be of n b = 1 ans = x # iterates from right # to left in a digit while (x!=0) : # while iterating this # is the number from # right to left cur = (x - 1) * b + (b - 1) # calls the function to # check if sum of cur is # more then of ans if (sumOfDigits(cur) > sumOfDigits(ans) or (sumOfDigits(cur) == sumOfDigits(ans) and cur > ans)) : ans = cur # reduces the number # to one unit less x =x // 10 b = b * 10 return ans # driver program to test the above function n = 521 print(findMax(n)) # This article is contributed by Nikita Tiwari.

Output:

499

**Time complexity:** O(m) where m is the number of digits in n.

This article is contributed by **Striver**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.