Given two numbers, N and M, the task is to find two numbers made up of M as all its digits such that their difference is divisible by N.
Examples:
Input: N = 8, M = 2
Output: 22 222
Explanation: The difference between 222 and 22 (200) is divisible by 8Input: N = 17, M = 6
Output: 6 66666666666666666
Approach:
In this problem, we have to find numbers consisting of only one unique digit. Let’s say M is equal to 2, then we have to find A and B from numbers like 2, 22, 222, 2222 …so on. The difference between A and B should be divisible by N. For this condition to satisfy, we have to pick A and B such that the remainder of A and B when it is divided by N, is the same.
For a digit of length N+1 length, consisting of only one unique digit M, we will have N+1 numbers. If we divide these N+1 numbers by N we will have N+1 remainders which will range from [0, N]. Since the numbers can exceed the range of integer values, we are storing remainder-length of the number as key-value pairings in a Map. Once a remainder occurs with a value already paired in the Map, the current length and the mapped lengths are the lengths of the desired numbers.
The below code is the implementation of the above approach:
C++
// C++ implementation// of the above approach #include <bits/stdc++.h>using namespace std; // Function to implement// the above approachvoid findNumbers(int N, int M){ int m = M; // Hashmap to store // remainder-length of the // number as key-value pairs map<int, int> remLen; int len, remainder; // Iterate till N + 1 length for (len = 1; len <= N + 1; ++len) { remainder = M % N; // Search remainder in the map if (remLen.find(remainder) == remLen.end()) // If remainder is not // already present insert // the length for the // corresponding remainder remLen[remainder] = len; else break; // Keep increasing M M = M * 10 + m; // To keep M in range of integer M = M % N; } // Length of one number // is the current Length int LenA = len; // Length of the other number // is the length paired with // current remainder in map int LenB = remLen[remainder]; for (int i = 0; i < LenB; ++i) cout << m; cout << " "; for (int i = 0; i < LenA; ++i) cout << m; return;} // Driver codeint main(){ int N = 8, M = 2; findNumbers(N, M); return 0;} |
Java
// Java implementation of the above approachimport java.util.*; class GFG{ // Function to implement// the above approachstatic void findNumbers(int N, int M){ int m = M; // Hashmap to store // remainder-length of the // number as key-value pairs Map<Integer, Integer> remLen = new HashMap<>(); int len, remainder = 0; // Iterate till N + 1 length for(len = 1; len <= N + 1; ++len) { remainder = M % N; // Search remainder in the map if (!remLen.containsKey(remainder)) { // If remainder is not // already present insert // the length for the // corresponding remainder remLen.put(remainder, len); } else { break; } // Keep increasing M M = M * 10 + m; // To keep M in range of integer M = M % N; } // Length of one number // is the current Length int LenA = len; // Length of the other number // is the length paired with // current remainder in map int LenB = remLen.getOrDefault(remainder, 0); for(int i = 0; i < LenB; ++i) System.out.print(m); System.out.print(" "); for(int i = 0; i < LenA; ++i) System.out.print(m);} // Driver codepublic static void main(String[] args){ int N = 8, M = 2; findNumbers(N, M);}} // This code is contributed by offbeat |
Python3
# Python3 implementation# of the above approach # Function to implement# the above approachdef findNumbers(N, M): m = M # Hashmap to store # remainder-length of the # number as key-value pairs remLen = {} # Iterate till N + 1 length for len1 in range(1, N + 1, 1): remainder = M % N # Search remainder in the map if (remLen.get(remainder) == None): # If remainder is not # already present insert # the length for the # corresponding remainder remLen[remainder] = len1 else: break # Keep increasing M M = M * 10 + m # To keep M in range of integer M = M % N # Length of one number # is the current Length LenA = len1 # Length of the other number # is the length paired with # current remainder in map LenB = remLen[remainder] for i in range(LenB): print(m, end = "") print(" ", end = "") for i in range(LenA): print(m, end = "") return # Driver codeif __name__ == '__main__': N = 8 M = 2 findNumbers(N, M) # This code is contributed by Bhupendra_Singh |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic; class GFG{ // Function to implement// the above approachstatic void findNumbers(int N, int M){ int m = M; // To store remainder-length of // the number as key-value pairs Dictionary<int, int> remLen = new Dictionary<int, int>(); int len, remainder = 0; // Iterate till N + 1 length for(len = 1; len <= N + 1; ++len) { remainder = M % N; // Search remainder in the map if (!remLen.ContainsKey(remainder)) { // If remainder is not // already present insert // the length for the // corresponding remainder remLen.Add(remainder, len); } else { break; } // Keep increasing M M = M * 10 + m; // To keep M in range of integer M = M % N; } // Length of one number // is the current Length int LenA = len; // Length of the other number // is the length paired with // current remainder in map int LenB = remLen[remainder]; for(int i = 0; i < LenB; ++i) Console.Write(m); Console.Write(" "); for(int i = 0; i < LenA; ++i) Console.Write(m);} // Driver codepublic static void Main(String[] args){ int N = 8, M = 2; findNumbers(N, M);}} // This code is contributed by Amit Katiyar |
22 222
Time Complexity: O(N)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.
In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA
