Given a number
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++ program to find the count of Steps to // reduce N to zero by subtracting its most // significant digit at every step #include <bits/stdc++.h> 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 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 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# 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 // PHP 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 function countdig( $m )
{ if ( $m == 0)
return 0;
else
return 1 + countdig( (int)( $m / 10));
} // Function to count the number of // steps to reach 0 function 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 = (int)( $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;
echo countSteps( $n );
// This code is contributed // by Akanksha Rai |
<script> // javascript 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
function countdig(m) {
if (m == 0)
return 0;
else
return 1 + countdig(parseInt(m / 10));
}
// Function to count the number of
// steps to reach 0
function countSteps(x) {
// count the total number of steps
var c =0;
var last = x;
// iterate till we reach 0
while (last > 0) {
// count the digits in last
var digits = countdig(last);
// decrease it by 1
digits -= 1;
// find the number on whose division,
// we get the first digit
var divisor = parseInt( Math.pow(10, digits));
// first digit in last
var first = parseInt(last / divisor);
// find the first number less than
// last where the first digit changes
var lastnumber = first * divisor;
// find the number of numbers
// with same first digit that are jumped
var skipped = parseInt((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
var n = 14;
document.write(countSteps(n));
// This code is contributed by todaysgaurav </script> |
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.