Given two integers N and K, the task is to print the number formed by adding product of its max and min digit, K times. 
Examples
Input: N = 14, K = 3
Output: 26
Explanation:
M(0)=14
M(1)=14 + 1*4 = 18
M(2)=18 + 1*8 = 26Input: N = 487, K = 100000000
Output: 950
Approach
- A natural intuition is to run a loop K times and keep updating the value of N.
- But one observation can be observed that after some iteration minimum value of digit may be zero and after that N is never going to be updated because:
M(N + 1) = M(N) + 0*(max_digit)
M(N + 1) = M(N)
- Hence we just need to figure out when minimum digit became 0.
Below is the implementation of the above approach:
C++
// C++ Code for the above approach#include <bits/stdc++.h>using namespace std;// function that returns the product of// maximum and minimum digit of N number.int prod_of_max_min(int n){ int largest = 0; int smallest = 10; while (n) { // finds the last digit. int r = n % 10; largest = max(r, largest); smallest = min(r, smallest); // Moves to next digit n = n / 10; } return largest * smallest;}// Function to find the formed numberint formed_no(int N, int K){ if (K == 1) { return N; } K--; // M(1) = N int answer = N; while (K--) { int a_current = prod_of_max_min(answer); // check if minimum digit is 0 if (a_current == 0) break; answer += a_current; } return answer;}// Driver Codeint main(){ int N = 487, K = 100000000; cout << formed_no(N, K) << endl; return 0;} |
Java
// Java program for the above approachimport java.io.*; import java.util.*;class GFG { // Function to find the formed number public static int formed_no(int N, int K){ if (K == 1) { return N; } K--; // M(1) = N int answer = N; while(K != 0) { int a_current = prod_of_max_min(answer); // Check if minimum digit is 0 if (a_current == 0) break; answer += a_current; } return answer;}// Function that returns the product of // maximum and minimum digit of N number.static int prod_of_max_min(int n){ int largest = 0; int smallest = 10; while(n != 0) { // Finds the last digit. int r = n % 10; largest = Math.max(r, largest); smallest = Math.min(r, smallest); // Moves to next digit n = n / 10; } return largest * smallest;} // Driver code public static void main(String[] args) { int N = 487, K = 100000000; System.out.println(formed_no(N, K));} } // This code is contributed by coder001 |
Python3
# Python3 program for the above approach# Function to find the formed numberdef formed_no(N, K): if (K == 1): return N K -= 1 # M(1) = N answer = N while (K != 0): a_current = prod_of_max_min(answer) # Check if minimum digit is 0 if (a_current == 0): break answer += a_current K -= 1 return answer# Function that returns the product of# maximum and minimum digit of N number.def prod_of_max_min(n): largest = 0 smallest = 10 while (n != 0): # Find the last digit. r = n % 10 largest = max(r, largest) smallest = min(r, smallest) # Moves to next digit n = n // 10 return largest * smallest# Driver Codeif __name__ == "__main__": N = 487 K = 100000000 print(formed_no(N, K))# This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;class GFG { // Function to find the formed number public static int formed_no(int N, int K){ if (K == 1) { return N; } K--; // M(1) = N int answer = N; while(K != 0) { int a_current = prod_of_max_min(answer); // Check if minimum digit is 0 if (a_current == 0) break; answer += a_current; } return answer;}// Function that returns the product of // maximum and minimum digit of N number.static int prod_of_max_min(int n){ int largest = 0; int smallest = 10; while(n != 0) { // Finds the last digit. int r = n % 10; largest = Math.Max(r, largest); smallest = Math.Min(r, smallest); // Moves to next digit n = n / 10; } return largest * smallest;} // Driver code public static void Main(String[] args) { int N = 487, K = 100000000; Console.WriteLine(formed_no(N, K));} } // This code is contributed by Rohit_ranjan |
Javascript
<script>// JavaScript Code for the above approach// Function that returns the product of// maximum and minimum digit of N number.function prod_of_max_min(n) { var largest = 0; var smallest = 10; while (n) { // Finds the last digit. var r = n % 10; largest = Math.max(r, largest); smallest = Math.min(r, smallest); // Moves to next digit n = parseInt(n / 10); } return largest * smallest;}// Function to find the formed numberfunction formed_no(N, K) { if (K == 1) { return N; } K--; // M(1) = N var answer = N; while (K--) { var a_current = prod_of_max_min(answer); // Check if minimum digit is 0 if (a_current == 0) break; answer += a_current; } return answer;}// Driver Codevar N = 487,K = 100000000;document.write(formed_no(N, K) + "<br>");// This code is contributed by rdtank</script> |
Output:
950
Time Complexity: O(K)
Auxiliary Space: O(1)
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