Largest number with maximum trailing nines which is less than N and greater than N-D
Given two numbers N and D. The task is to find out the largest number smaller than or equal to N which contains the maximum number of trailing nines and the difference between N and the number should not be greater than D.
Examples:
Input: N = 1029, D = 102 Output: 999 1029 has 1 trailing nine while 999 has three trailing nine.Also 1029-999 = 30(which is less than 102). Input: N = 10281, D = 1 Output: 10281
A naive approach will be to iterate from N till N-D and find the number with the largest number of trailing nines.
An efficient approach can be found by some key observations. One key observation for this problem is that the largest number smaller than N ending with at least say(K) nines is
[n – (n MOD 10^k) – 1]
Traverse all possible values of k starting from total no of digits of N to 1, and check whether d > n%
. If no such value is obtained, the final answer will be N itself. Otherwise, check for the answer using the above observation.
Below is the implementation of the above approach.
C++
// CPP to implement above function #include <bits/stdc++.h> using namespace std; // It's better to use long long // to handle big integers #define ll long long // function to count no of digits ll dig(ll a) { ll count = 0; while (a > 0) { a /= 10; count++; } return count; } // function to implement above approach void required_number(ll num, ll n, ll d) { ll i, j, power, a, flag = 0; for (i = num; i >= 1; i--) { power = pow(10, i); a = n % power; // if difference between power // and n doesn't exceed d if (d > a) { flag = 1; break; } } if (flag) { ll t = 0; // loop to build a number from the // appropriate no of digits containg only 9 for (j = 0; j < i; j++) { t += 9 * pow(10, j); } // if the build number is // same as original number(n) if (n % power == t) cout << n; else { // observation cout << n - (n % power) - 1; } } else cout << n; } // Driver Code int main() { ll n = 1029, d = 102; // variable that stores no of digits in n ll num = dig(n); required_number(num, n, d); return 0; } |
Java
// Java code to implement above function import java.io.*; class GFG { // It's better to use long // to handle big integers // function to count no. of digits static long dig(long a) { long count = 0; while (a > 0) { a /= 10; count++; } return count; } // function to implement above approach static void required_number(long num, long n, long d) { long i, j, power=1, a, flag = 0; for (i = num; i >= 1; i--) { power = (long)Math.pow(10, i); a = n % power; // if difference between power // and n doesn't exceed d if (d > a) { flag = 1; break; } } if (flag>0) { long t = 0; // loop to build a number from the // appropriate no of digits containg // only 9 for (j = 0; j < i; j++) { t += 9 * Math.pow(10, j); } // if the build number is // same as original number(n) if (n % power == t) System.out.print( n); else { // observation System.out.print( n - (n % power) - 1); } } else System.out.print(n); } // Driver Code public static void main (String[] args) { long n = 1029, d = 102; // variable that stores no // of digits in n long num = dig(n); required_number(num, n, d); } } // This code is contributed by chandan_jnu |
Python3
# Python3 to implement above function # function to count no of digits def dig(a): count = 0; while (a > 0): a /= 10 count+=1 return count # function to implement above approach def required_number(num, n, d): flag = 0 power=0 a=0 for i in range(num,0,-1): power = pow(10, i) a = n % power # if difference between power # and n doesn't exceed d if (d > a): flag = 1 break if(flag): t=0 # loop to build a number from the # appropriate no of digits containg only 9 for j in range(0,i): t += 9 * pow(10, j) # if the build number is # same as original number(n) if(n % power ==t): print(n,end="") else: # observation print((n - (n % power) - 1),end="") else: print(n,end="") # Driver Code if __name__ == "__main__": n = 1029 d = 102 # variable that stores no of digits in n num = dig(n) required_number(num, n, d) # this code is contributed by mits |
C#
// C# code to implement // above function using System; class GFG { // It's better to use long // to handle big integers // function to count no. of digits static long dig(long a) { long count = 0; while (a > 0) { a /= 10; count++; } return count; } // function to implement // above approach static void required_number(long num, long n, long d) { long i, j, power = 1, a, flag = 0; for (i = num; i >= 1; i--) { power = (long)Math.Pow(10, i); a = n % power; // if difference between power // and n doesn't exceed d if (d > a) { flag = 1; break; } } if (flag > 0) { long t = 0; // loop to build a number // from the appropriate no // of digits containg only 9 for (j = 0; j < i; j++) { t += (long)(9 * Math.Pow(10, j)); } // if the build number is // same as original number(n) if (n % power == t) Console.Write( n); else { // observation Console.Write(n - (n % power) - 1); } } else Console.Write(n); } // Driver Code public static void Main() { long n = 1029, d = 102; // variable that stores // no. of digits in n long num = dig(n); required_number(num, n, d); } } // This code is contributed // by chandan_jnu |
PHP
<?php // PHP to implement above function // function to count no of digits function dig($a) { $count = 0; while ($a > 0) { $a = (int)($a / 10); $count++; } return $count; } // function to implement above approach function required_number($num, $n, $d) { $flag = 0; for ($i = $num; $i >= 1; $i--) { $power = pow(10, $i); $a = $n % $power; // if difference between power // and n doesn't exceed d if ($d > $a) { $flag = 1; break; } } if ($flag) { $t = 0; // loop to build a number from the // appropriate no of digits containg only 9 for ($j = 0; $j < $i; $j++) { $t += 9 * pow(10, $j); } // if the build number is // same as original number(n) if ($n % $power == $t) echo $n; else { // observation echo ($n - ($n % $power) - 1); } } else echo $n; } // Driver Code $n = 1029; $d = 102; // variable that stores no of // digits in n $num = dig($n); required_number($num, $n, $d); // This code is contributed by mits ?> |
Output:
999
Time Complexity:O(no of digits)
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