# Minimum numbers with one’s place as 9 to be added to get N

Given an integer N, the task is to find the minimum count of numbers which needs to be added to get N. (All such numbers must have 9 as one’s digit)

Examples:

Input: N = 27
Output: 3
27 = 9 + 9 + 9

Input: N = 109
Output: 1
109 itself has 9 as one’s digit.

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

Approach:

• Check the one’s digit of N, based on one’s digit the minimum count of numbers which needs to be added can be easily found.
• If one’s digit is 9: The answer will 1 as the number itself has 9 as its one’s place digit.
• If one’s digit is:
1. 1: 9 has to be added 9 times i.e. (9 * 9 = 81).
2. 2: 9 has to be added 8 times i.e. (9 * 8 = 72).
3. 3: 9 has to be added 7 times i.e. (9 * 7 = 63).
4. 4: 9 has to be added 6 times i.e. (9 * 6 = 54).
5. 5: 9 has to be added 5 times i.e. (9 * 5 = 45).
6. 6: 9 has to be added 4 times i.e. (9 * 4 = 36).
7. 7: 9 has to be added 3 times i.e. (9 * 3 = 27).
8. 8: 9 has to be added 2 times i.e. (9 * 2 = 18).
9. 0: 9 has to be added 10 times i.e. (9 * 10 = 90).
• Observation here is that only the minimum multiple count for all the cases above mentioned have to be added. This is because say for one’s digit as 4, all 9 (one’s place) from 6 numbers can be used and the result can be subtracted from N say it is M. Now, M will have 0 in one’s place. So, just use 5 numbers as 9 and sixth number as (M + 9).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find minimum count ` `// of numbers(with one's digit 9) ` `// that sum up to N ` `int` `findMin(``int` `N) ` `{ ` `    ``// Fetch one's digit ` `    ``int` `digit = N % 10; ` ` `  `    ``// Apply Cases mentioned in approach ` `    ``switch` `(digit) { ` `    ``case` `0: ` `        ``if` `(N >= 90) ` `            ``return` `10; ` `        ``break``; ` `    ``case` `1: ` `        ``if` `(N >= 81) ` `            ``return` `9; ` `        ``break``; ` `    ``case` `2: ` `        ``if` `(N >= 72) ` `            ``return` `8; ` `        ``break``; ` `    ``case` `3: ` `        ``if` `(N >= 63) ` `            ``return` `7; ` `        ``break``; ` `    ``case` `4: ` `        ``if` `(N >= 54) ` `            ``return` `6; ` `        ``break``; ` `    ``case` `5: ` `        ``if` `(N >= 45) ` `            ``return` `5; ` `        ``break``; ` `    ``case` `6: ` `        ``if` `(N >= 36) ` `            ``return` `4; ` `        ``break``; ` `    ``case` `7: ` `        ``if` `(N >= 27) ` `            ``return` `3; ` `        ``break``; ` `    ``case` `8: ` `        ``if` `(N >= 18) ` `            ``return` `2; ` `        ``break``; ` `    ``case` `9: ` `        ``if` `(N >= 9) ` `            ``return` `1; ` `        ``break``; ` `    ``} ` ` `  `    ``// If no possible answer exists ` `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 27; ` ` `  `    ``cout << findMin(N); ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `    ``// Function to find minimum count  ` `    ``// of numbers(with one's digit 9)  ` `    ``// that sum up to N  ` `    ``static` `int` `findMin(``int` `N)  ` `    ``{  ` `        ``// Fetch one's digit  ` `        ``int` `digit = N % ``10``;  ` `     `  `        ``// Apply Cases mentioned in approach  ` `        ``switch` `(digit) ` `        ``{  ` `            ``case` `0``:  ` `                ``if` `(N >= ``90``)  ` `                    ``return` `10``;  ` `                ``break``;  ` `            ``case` `1``:  ` `                ``if` `(N >= ``81``)  ` `                    ``return` `9``;  ` `                ``break``;  ` `            ``case` `2``:  ` `                ``if` `(N >= ``72``)  ` `                    ``return` `8``;  ` `                ``break``;  ` `            ``case` `3``:  ` `                ``if` `(N >= ``63``)  ` `                    ``return` `7``;  ` `                ``break``;  ` `            ``case` `4``:  ` `                ``if` `(N >= ``54``)  ` `                    ``return` `6``;  ` `                ``break``;  ` `            ``case` `5``:  ` `                ``if` `(N >= ``45``)  ` `                    ``return` `5``;  ` `                ``break``;  ` `            ``case` `6``:  ` `                ``if` `(N >= ``36``)  ` `                    ``return` `4``;  ` `                ``break``;  ` `            ``case` `7``:  ` `                ``if` `(N >= ``27``)  ` `                    ``return` `3``;  ` `                ``break``;  ` `            ``case` `8``:  ` `                ``if` `(N >= ``18``)  ` `                    ``return` `2``;  ` `                ``break``;  ` `            ``case` `9``:  ` `                ``if` `(N >= ``9``)  ` `                    ``return` `1``;  ` `                ``break``;  ` `        ``}  ` `     `  `        ``// If no possible answer exists  ` `        ``return` `-``1``;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `N = ``27``;  ` `     `  `        ``System.out.println(findMin(N));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to find minimum count ` `# of numbers(with one's digit 9) ` `# that sum up to N ` `def` `findMin(N: ``int``): ` ` `  `    ``# Fetch one's digit ` `    ``digit ``=` `N ``%` `10` ` `  `    ``# Apply Cases mentioned in approach ` `    ``if` `digit ``=``=` `0` `and` `N >``=` `90``: ` `        ``return` `10` `    ``elif` `digit ``=``=` `1` `and` `N >``=` `81``: ` `        ``return` `9` `    ``elif` `digit ``=``=` `2` `and` `N >``=` `72``: ` `        ``return` `8` `    ``elif` `digit ``=``=` `3` `and` `N >``=` `63``: ` `        ``return` `7` `    ``elif` `digit ``=``=` `4` `and` `N >``=` `54``: ` `        ``return` `6` `    ``elif` `digit ``=``=` `5` `and` `N >``=` `45``: ` `        ``return` `5` `    ``elif` `digit ``=``=` `6` `and` `N >``=` `36``: ` `        ``return` `4` `    ``elif` `digit ``=``=` `7` `and` `N >``=` `27``: ` `        ``return` `3` `    ``elif` `digit ``=``=` `8` `and` `N >``=` `18``: ` `        ``return` `2` `    ``elif` `digit ``=``=` `9` `and` `N >``=` `9``: ` `        ``return` `1` ` `  `    ``# If no possible answer exists ` `    ``return` `-``1` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``N ``=` `27` `    ``print``(findMin(N)) ` ` `  `# This code is contributed by ` `# sanjeev2552 `

## C#

 `// C# implementation of the approach  ` `using` `System;          ` `     `  `class` `GFG ` `{ ` `     `  `    ``// Function to find minimum count  ` `    ``// of numbers(with one's digit 9)  ` `    ``// that sum up to N  ` `    ``static` `int` `findMin(``int` `N)  ` `    ``{  ` `        ``// Fetch one's digit  ` `        ``int` `digit = N % 10;  ` `     `  `        ``// Apply Cases mentioned in approach  ` `        ``switch` `(digit) ` `        ``{  ` `            ``case` `0:  ` `                ``if` `(N >= 90)  ` `                    ``return` `10;  ` `                ``break``;  ` `            ``case` `1:  ` `                ``if` `(N >= 81)  ` `                    ``return` `9;  ` `                ``break``;  ` `            ``case` `2:  ` `                ``if` `(N >= 72)  ` `                    ``return` `8;  ` `                ``break``;  ` `            ``case` `3:  ` `                ``if` `(N >= 63)  ` `                    ``return` `7;  ` `                ``break``;  ` `            ``case` `4:  ` `                ``if` `(N >= 54)  ` `                    ``return` `6;  ` `                ``break``;  ` `            ``case` `5:  ` `                ``if` `(N >= 45)  ` `                    ``return` `5;  ` `                ``break``;  ` `            ``case` `6:  ` `                ``if` `(N >= 36)  ` `                    ``return` `4;  ` `                ``break``;  ` `            ``case` `7:  ` `                ``if` `(N >= 27)  ` `                    ``return` `3;  ` `                ``break``;  ` `            ``case` `8:  ` `                ``if` `(N >= 18)  ` `                    ``return` `2;  ` `                ``break``;  ` `            ``case` `9:  ` `                ``if` `(N >= 9)  ` `                    ``return` `1;  ` `                ``break``;  ` `        ``}  ` `     `  `        ``// If no possible answer exists  ` `        ``return` `-1;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args) ` `    ``{  ` `        ``int` `N = 27;  ` `     `  `        ``Console.WriteLine(findMin(N));  ` `    ``}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```3
```

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