Given an array arr[] of size N and an integer K. The task is to find the largest K digit number divisible by all number of arr[].
Examples:
Input: arr[] = {2, 3, 5}, K = 3
Output: 990
Explanation: 990 is the largest 3 digit number divisible by 2, 3 and 5.Input: arr[] = {91, 93, 95}, K = 3
Output: -1
Explanation: There is no 3 digit number divisible by all 91, 93 and 95.
Approach: The solution is based on the idea similar to finding largest K digit number divisible by X. Follow the steps mentioned below.
- Find LCM of all numbers of array arr[]
- Find the largest multiple of LCM having K digits using the formula:
LCM(arr) * ((10K-1)/LCM(arr))
Below is the implementation of the above approach.
C++
// C++ code to implement above approach #include <bits/stdc++.h>using namespace std; int __gcd(int a, int b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // Base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Function to find LCM of the array int findLCM(int arr[], int n, int idx) { // lcm(a, b) = (a*b / gcd(a, b)) if (idx == n - 1) { return arr[idx]; } int a = arr[idx]; int b = findLCM(arr, n, idx + 1); double gcd = __gcd(a, b); return (a * (int)floor(b / gcd)); } // Function to find the number int findNum(int arr[], int n, int K) { int lcm = findLCM(arr, n, 0); int ans = (int)floor((pow(10, K) - 1) / lcm); ans = (ans) * lcm; if (ans == 0) return -1; return ans; } // Driver code int main() { int arr[] = { 2, 3, 5 }; int n = sizeof(arr) / sizeof(arr[0]); int K = 3; cout << findNum(arr, n, K); }// This code is contributed by Samim Hossain Mondal. |
Java
// Java code to implement above approach class GFG{ static int __gcd(int a, int b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // Base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Function to find LCM of the array static int findLCM(int []arr, int idx) { // lcm(a, b) = (a*b / gcd(a, b)) if (idx == arr.length - 1) { return arr[idx]; } int a = arr[idx]; int b = findLCM(arr, idx + 1); double gcd = __gcd(a, b); return (a * (int)Math.floor(b / gcd)); } // Function to find the number static int findNum(int []arr, int K) { int lcm = findLCM(arr, 0); int ans = (int)Math.floor((Math.pow(10, K) - 1) / lcm); ans = (ans) * lcm; if (ans == 0) return -1; return ans; } // Driver code public static void main(String []args) { int []arr = { 2, 3, 5 }; int K = 3; System.out.println(findNum(arr, K)); }}// This code is contributed by 29AjayKumar |
Python3
# python code to implement above approachimport math# Function to find LCM of the arraydef findLCM(arr, idx): # lcm(a, b) = (a*b / gcd(a, b)) if (idx == len(arr) - 1): return arr[idx] a = arr[idx] b = findLCM(arr, idx + 1) return (a * b // math.gcd(a, b))# Function to find the numberdef findNum(arr, K): lcm = findLCM(arr, 0) ans = (pow(10, K) - 1) // lcm ans = (ans)*lcm if (ans == 0): return -1 return ans# Driver codeif __name__ == "__main__": arr = [2, 3, 5] K = 3 print(findNum(arr, K))# This code is contributed by rakeshsahni |
C#
// C# code to implement above approach using System;using System.Collections;class GFG{static int __gcd(int a, int b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // Base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a);}// Function to find LCM of the arraystatic int findLCM(int []arr, int idx) { // lcm(a, b) = (a*b / gcd(a, b)) if (idx == arr.Length - 1) { return arr[idx]; } int a = arr[idx]; int b = findLCM(arr, idx + 1); double gcd = __gcd(a, b); return (a * (int)Math.Floor(b / gcd));}// Function to find the numberstatic int findNum(int []arr, int K){ int lcm = findLCM(arr, 0); int ans = (int)Math.Floor((Math.Pow(10, K) - 1) / lcm); ans = (ans) * lcm; if (ans == 0) return -1; return ans;}// Driver codepublic static void Main(){int []arr = { 2, 3, 5 };int K = 3;Console.Write(findNum(arr, K));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// JavaScript code to implement above approach function __gcd(a, b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // Base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a);}// Function to find LCM of the arrayfunction findLCM(arr, idx) { // lcm(a, b) = (a*b / gcd(a, b)) if (idx == arr.length - 1) { return arr[idx]; } let a = arr[idx]; let b = findLCM(arr, idx + 1); return Math.floor((a * b / __gcd(a, b)));}// Function to find the numberfunction findNum(arr, K){ let lcm = findLCM(arr, 0); let ans = Math.floor((Math.pow(10, K) - 1) / lcm); ans = (ans) * lcm; if (ans == 0) return -1; return ans;}// Driver codelet arr = [ 2, 3, 5 ];let K = 3;document.write(findNum(arr, K));// This code is contributed by Potta Lokesh</script> |
Output
990
Time Complexity: O(N*logD) where D is the maximum element of the array
Auxiliary Space: O(1)
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