Represent a number as the sum of positive numbers ending with 9

Given an integer N, the task is to check if N can be expressed as a sum of integers having 9 as the last digit (9, 19, 29, 39…), or not. If found to be true, then find the minimum count of such integers required to obtain N. Otherwise print -1.

Examples:

Input: N = 156
Output: 4
Explanation:
156 = 9 + 9 + 9 + 129

Input: N = 60
Output: -1
Explanation:
No possible way to obtain sum 60 from numbers having 9 as the last digit.

Naive Approach: This problem can be viewed as a variation of the Coin change problem. For this problem, the coins can be replaced with [9, 19, 29, 39…. up to the last number smaller than N that ends with 9].

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized based on the observation that if the last digit of N is K, then exactly (10 – K) minimum numbers are required to form N.

A sum N can be obtained by adding 10 – K numbers, where K is the last digit of N.
Therefore, sum N can be obtained by adding 9, (9 – K) times and adding N – (9 * (9 – K)) finally.

Follow the steps below to solve the problem:

1. Extract the last digit of the given number, K = N % 10
2. Using the above observation, a total of (10 – K) numbers are required. Now, calculate 9 * (9 – K), as the first 9 – K numbers required to obtain N is 9.
3. Now, calculate N – 9 * (9 – K) and store in a variable, say z. If z is greater than or equal to 9 and has 9 as its last digit, print 10 – K as the answer. Otherwise, print -1.

Below is the implementation of the above approach:

C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the minimum count` `// of numbers ending with 9 to form N` `int` `minCountOfNumbers(``int` `N)` `{` `    ``// Extract last digit of N` `    ``int` `k = N % 10;`   `    ``// Calculate the last digit` `    ``int` `z = N - (9 * (9 - k));`   `    ``// If the last digit` `    ``// satisfies the condition` `    ``if` `(z >= 9 && z % 10 == 9) {` `        ``return` `10 - k;` `    ``}` `    ``else` `        ``return` `-1;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 156;` `    ``cout << minCountOfNumbers(N);`   `    ``return` `0;` `}`

Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the minimum count` `// of numbers ending with 9 to form N` `static` `int` `minCountOfNumbers(``int` `N)` `{` `    `  `    ``// Extract last digit of N` `    ``int` `k = N % ``10``;`   `    ``// Calculate the last digit` `    ``int` `z = N - (``9` `* (``9` `- k));`   `    ``// If the last digit` `    ``// satisfies the condition` `    ``if` `(z >= ``9` `&& z % ``10` `== ``9``)` `    ``{` `        ``return` `10` `- k;` `    ``}` `    ``else` `        ``return` `-``1``;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``156``;` `    ``System.out.print(minCountOfNumbers(N));` `}` `}`   `// This code is contributed by 29AjayKumar `

Python3

 `# Python3 program for the above approach`   `# Function to find the minimum count` `# of numbers ending with 9 to form N` `def` `minCountOfNumbers(N):` `    `  `    ``# Extract last digit of N` `    ``k ``=` `N ``%` `10`   `    ``# Calculate the last digit` `    ``z ``=` `N ``-` `(``9` `*` `(``9` `-` `k))`   `    ``# If the last digit` `    ``# satisfies the condition` `    ``if` `(z >``=` `9` `and` `z ``%` `10` `=``=` `9``):` `        ``return` `10` `-` `k` `    ``else``:` `        ``return` `-``1`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``N ``=` `156` `    `  `    ``print``(minCountOfNumbers(N))`   `# This code is contributed by mohit kumar 29`

C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to find the minimum count` `// of numbers ending with 9 to form N` `static` `int` `minCountOfNumbers(``int` `N)` `{` `    `  `    ``// Extract last digit of N` `    ``int` `k = N % 10;`   `    ``// Calculate the last digit` `    ``int` `z = N - (9 * (9 - k));`   `    ``// If the last digit` `    ``// satisfies the condition` `    ``if` `(z >= 9 && z % 10 == 9)` `    ``{` `        ``return` `10 - k;` `    ``}` `    ``else` `        ``return` `-1;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `N = 156;` `    `  `    ``Console.Write(minCountOfNumbers(N));` `}` `}`   `// This code is contributed by 29AjayKumar`

Javascript

 ``

Output:

`4`

Time Complexity: O(1)
Auxiliary Space: O(1)

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