# Biggest integer which has maximum digit sum in range from 1 to n

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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. `

## C#

 `// C# program to find the number ` `// with maximum digit sum. ` `using` `System; ` ` `  `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() ` `    ``{ ` `        ``int` `n = 521;  ` `        ``Console.WriteLine(findMax(n)); ` `    ``} ` `} ` `  `  `// This article is contributed by Anant Agarwal. `

## PHP

 ` sumOfDigits(``\$ans``) ||  ` `           ``(sumOfDigits(``\$cur``) == sumOfDigits(``\$ans``) &&  ` `                                        ``\$cur` `> ``\$ans``)) ` `            ``\$ans` `= ``\$cur``; ` ` `  `        ``// reduces the number ` `        ``// to one unit less ` `        ``\$x` `= (int)``\$x` `/ 10; ` `        ``\$b` `*= 10; ` `    ``} ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `// Driver Code ` `\$n` `= 521;  ` `echo` `findMax(``\$n``);  ` ` `  `// This code is contributed by ajit ` `?> `

Output :

```499
```

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

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