Given the integers a, b, N, the task is to find the largest number x such that 
is an N-digit number of base b.
Examples:
Input: a = 2, b = 10, N = 2
Output: 3
Explanation:
Here 2 * 33 = 54, which has the number of digits = 2,
but 2 * 44 = 512 which has a number of digits = 3, which is not equal to N.
Therefore, the largest value of x is 2.Input: a = 1, b = 2, N = 3
Output: 2
Explanation:
1 * 22 = 4 whose binary representation is 100 and it has 3 digits.
Approach: This problem can be solved using binary search.
- The number of digits of
in base 
is 
. - Binary search is used to find the largest
such that the number of digits of in base is exactly 
. - In binary search, we will check the number of digits , where
, and change the pointer according to that. 
in base 
is 
.
such that the number of digits of 
.
, where 
, and change the pointer according to that. 
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find log_b(a)double log(int a, int b){ return log10(a) / log10(b);}int get(int a, int b, int n){ // Set two pointer for binary search int lo = 0, hi = 1e6; int ans = 0; while (lo <= hi) { int mid = (lo + hi) / 2; // Calculating number of digits // of a*mid^mid in base b int dig = ceil((mid * log(mid, b) + log(a, b))); if (dig > n) { // If number of digits > n // we can simply ignore it // and decrease our pointer hi = mid - 1; } else { // if number of digits <= n, // we can go higher to // reach value exactly equal to n ans = mid; lo = mid + 1; } } // return the largest value of x return ans;}// Driver Codeint main(){ int a = 2, b = 2, n = 6; cout << get(a, b, n) << "\n"; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{// Function to find log_b(a)static int log(int a, int b){ return (int)(Math.log10(a) / Math.log10(b));}static int get(int a, int b, int n){ // Set two pointer for binary search int lo = 0, hi = (int) 1e6; int ans = 0; while (lo <= hi) { int mid = (lo + hi) / 2; // Calculating number of digits // of a*mid^mid in base b int dig = (int) Math.ceil((mid * log(mid, b) + log(a, b))); if (dig > n) { // If number of digits > n // we can simply ignore it // and decrease our pointer hi = mid - 1; } else { // If number of digits <= n, // we can go higher to reach // value exactly equal to n ans = mid; lo = mid + 1; } } // Return the largest value of x return ans;}// Driver Codepublic static void main(String[] args){ int a = 2, b = 2, n = 6; System.out.print(get(a, b, n) + "\n");}}// This code is contributed by amal kumar choubey |
Python3
# Python3 implementation of the above approachfrom math import log10,ceil,log# Function to find log_b(a)def log1(a,b): return log10(a)//log10(b)def get(a,b,n): # Set two pointer for binary search lo = 0 hi = 1e6 ans = 0 while (lo <= hi): mid = (lo + hi) // 2 # Calculating number of digits # of a*mid^mid in base b dig = ceil((mid * log(mid, b) + log(a, b))) if (dig > n): # If number of digits > n # we can simply ignore it # and decrease our pointer hi = mid - 1 else: # if number of digits <= n, # we can go higher to # reach value exactly equal to n ans = mid lo = mid + 1 # return the largest value of x return ans# Driver Codeif __name__ == '__main__': a = 2 b = 2 n = 6 print(int(get(a, b, n)))# This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the above approachusing System;class GFG{// Function to find log_b(a)static int log(int a, int b){ return (int)(Math.Log10(a) / Math.Log10(b));}static int get(int a, int b, int n){ // Set two pointer for binary search int lo = 0, hi = (int) 1e6; int ans = 0; while (lo <= hi) { int mid = (lo + hi) / 2; // Calculating number of digits // of a*mid^mid in base b int dig = (int)Math.Ceiling((double)(mid * log(mid, b) + log(a, b))); if (dig > n) { // If number of digits > n // we can simply ignore it // and decrease our pointer hi = mid - 1; } else { // If number of digits <= n, // we can go higher to reach // value exactly equal to n ans = mid; lo = mid + 1; } } // Return the largest value of x return ans;}// Driver Codepublic static void Main(String[] args){ int a = 2, b = 2, n = 6; Console.Write(get(a, b, n) + "\n");}}// This code is contributed by amal kumar choubey |
Javascript
<script>// Javascript implementation of the above approach// Function to find log_b(a)function log(a, b){ return (Math.log10(a) / Math.log10(b));} function get(a, b, n){ // Set two pointer for binary search let lo = 0, hi = 1e6; let ans = 0; while (lo <= hi) { let mid = Math.floor((lo + hi) / 2); // Calculating number of digits // of a*mid^mid in base b let dig = Math.ceil((mid * log(mid, b) + log(a, b))); if (dig > n) { // If number of digits > n // we can simply ignore it // and decrease our pointer hi = mid - 1; } else { // If number of digits <= n, // we can go higher to reach // value exactly equal to n ans = mid; lo = mid + 1; } } // Return the largest value of x return ans;}// Driver Code let a = 2, b = 2, n = 6; document.write(get(a, b, n) + "\n"); </script> |
Output:
3
Time Complexity: O(log(k)) where k=1e6.
Auxiliary Space: O(1)
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