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# Steps to reduce N to zero by subtracting its most significant digit at every step

• Difficulty Level : Basic
• Last Updated : 13 Jun, 2022

Given a number . Reduce this number to zero by subtracting the number by it’s most significant digit(Left most digit) at every step. The task is to count the number of steps it takes to be reduced to zero.
Examples

```Input: 14
Output: 6
Steps:
14 - 1 = 13
13 - 1 = 12
12 - 1 = 11
11 - 1 = 10
10 - 1 = 9
9 - 9 = 0

Input: 20
Output: 12
Numbers after series of steps:
20, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 0```

Naive Approach: A naive approach is to reduce the number by its first digit step-wise and find the count of steps, but the time complexity will be huge if a large number is provided.
Efficient Approach: The main idea of the efficient approach is to reduce the number of steps in the naive approach. We can skip the steps whose leading digits are the same in consecutive numbers, and count them. The algorithm of skipping those numbers with the same leading digits is as follows:

• Let the number be last, count the digits in last and reduce it by 1, because the smallest number with same leading digit with the same count of digits will have that number of zeros in it.
• Find the first digit of the number of last, by last/count.
• Hence the smallest number of same number of count of digits with same leading number will be [first digit * (count-1)]
• the number of steps skipped can be achieved by [(last-smallest number)/first digit].
• Hence the next number last will be last – (first*skipped)

Below is the implementation of the above approach:

## C++

 `// C++ program to find the count of Steps to``// reduce N to zero by subtracting its most``// significant digit at every step` `#include ``using` `namespace` `std;` `// Function to count the number``// of digits in a number m``int` `countdig(``int` `m)``{``    ``if` `(m == 0)``        ``return` `0;``    ``else``        ``return` `1 + countdig(m / 10);``}` `// Function to count the number of``// steps to reach 0``int` `countSteps(``int` `x)``{``    ``// count the total number of steps``    ``int` `c = 0;``    ``int` `last = x;` `    ``// iterate till we reach 0``    ``while` `(last) {` `        ``// count the digits in last``        ``int` `digits = countdig(last);` `        ``// decrease it by 1``        ``digits -= 1;` `        ``// find the number on whose division,``        ``// we get the first digit``        ``int` `divisor = ``pow``(10, digits);` `        ``// first digit in last``        ``int` `first = last / divisor;` `        ``// find the first number less than``        ``// last where the first digit changes``        ``int` `lastnumber = first * divisor;` `        ``// find the number of numbers``        ``// with same first digit that are jumped``        ``int` `skipped = (last - lastnumber) / first;` `        ``skipped += 1;` `        ``// count the steps``        ``c += skipped;` `        ``// the next number with a different``        ``// first digit``        ``last = last - (first * skipped);``    ``}` `    ``return` `c;``}` `// Driver code``int` `main()``{``    ``int` `n = 14;` `    ``cout << countSteps(n);` `    ``return` `0;``}`

## Java

 `// Java program to find the count of Steps to``// reduce N to zero by subtracting its most``// significant digit at every step`  `class` `GFG{``// Function to count the number``// of digits in a number m``static` `int` `countdig(``int` `m)``{``    ``if` `(m == ``0``)``        ``return` `0``;``    ``else``        ``return` `1` `+ countdig(m / ``10``);``}` `// Function to count the number of``// steps to reach 0``static` `int` `countSteps(``int` `x)``{``    ``// count the total number of steps``    ``int` `c = ``0``;``    ``int` `last = x;` `    ``// iterate till we reach 0``    ``while` `(last>``0``) {` `        ``// count the digits in last``        ``int` `digits = countdig(last);` `        ``// decrease it by 1``        ``digits -= ``1``;` `        ``// find the number on whose division,``        ``// we get the first digit``        ``int` `divisor = (``int``)Math.pow(``10``, digits);` `        ``// first digit in last``        ``int` `first = last / divisor;` `        ``// find the first number less than``        ``// last where the first digit changes``        ``int` `lastnumber = first * divisor;` `        ``// find the number of numbers``        ``// with same first digit that are jumped``        ``int` `skipped = (last - lastnumber) / first;` `        ``skipped += ``1``;` `        ``// count the steps``        ``c += skipped;` `        ``// the next number with a different``        ``// first digit``        ``last = last - (first * skipped);``    ``}` `    ``return` `c;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``14``;` `    ``System.out.println(countSteps(n));``}``}``// This code is contributed by mits`

## Python 3

 `# Python 3 program to find the``# count of Steps to reduce N to``# zero by subtracting its most``# significant digit at every step` `# Function to count the number``# of digits in a number m``def` `countdig(m) :` `    ``if` `(m ``=``=` `0``) :``        ``return` `0``    ``else` `:``        ``return` `1` `+` `countdig(m ``/``/` `10``)` `# Function to count the number``# of steps to reach 0``def` `countSteps(x) :``    ` `    ``# count the total number``    ``# of steps``    ``c ``=` `0``    ``last ``=` `x` `    ``# iterate till we reach 0``    ``while` `(last) :` `        ``# count the digits in last``        ``digits ``=` `countdig(last)` `        ``# decrease it by 1``        ``digits ``-``=` `1` `        ``# find the number on whose``        ``# division, we get the first digit``        ``divisor ``=` `pow``(``10``, digits)` `        ``# first digit in last``        ``first ``=` `last ``/``/` `divisor` `        ``# find the first number less``        ``# than last where the first``        ``# digit changes``        ``lastnumber ``=` `first ``*` `divisor` `        ``# find the number of numbers``        ``# with same first digit that``        ``# are jumped``        ``skipped ``=` `(last ``-` `lastnumber) ``/``/` `first` `        ``skipped ``+``=` `1` `        ``# count the steps``        ``c ``+``=` `skipped` `        ``# the next number with a different``        ``# first digit``        ``last ``=` `last ``-` `(first ``*` `skipped)` `    ``return` `c` `# Driver code``n ``=` `14``print``(countSteps(n))` `# This code is contributed by ANKITRAI1`

## C#

 `// C# program to find the count of Steps to``// reduce N to zero by subtracting its most``// significant digit at every step``using` `System;` `class` `GFG{``// Function to count the number``// of digits in a number m``static` `int` `countdig(``int` `m)``{``    ``if` `(m == 0)``        ``return` `0;``    ``else``        ``return` `1 + countdig(m / 10);``}` `// Function to count the number of``// steps to reach 0``static` `int` `countSteps(``int` `x)``{``    ``// count the total number of steps``    ``int` `c = 0;``    ``int` `last = x;` `    ``// iterate till we reach 0``    ``while` `(last>0) {` `        ``// count the digits in last``        ``int` `digits = countdig(last);` `        ``// decrease it by 1``        ``digits -= 1;` `        ``// find the number on whose division,``        ``// we get the first digit``        ``int` `divisor = (``int``)Math.Pow(10, digits);` `        ``// first digit in last``        ``int` `first = last / divisor;` `        ``// find the first number less than``        ``// last where the first digit changes``        ``int` `lastnumber = first * divisor;` `        ``// find the number of numbers``        ``// with same first digit that are jumped``        ``int` `skipped = (last - lastnumber) / first;` `        ``skipped += 1;` `        ``// count the steps``        ``c += skipped;` `        ``// the next number with a different``        ``// first digit``        ``last = last - (first * skipped);``    ``}` `    ``return` `c;``}` `// Driver code``static` `void` `Main()``{``    ``int` `n = 14;` `    ``Console.WriteLine(countSteps(n));``}``}``// This code is contributed by mits`

## PHP

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## Javascript

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Output:

`6`

Time Complexity: O(N*logN), as in the worst case the while loop will traverse N times and in each traversal we are using power function and countdig function, each of these will cost logN time, so the effective time complexity of the program will be O (N*logN).

Auxiliary Space: O(1), as we are not using any extra space.

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