Given an array containing N elements. Each element is either 0 or 5. Find the largest number divisible by 90 that can be made using any number of elements of this array and arranging them in any way.
Examples:
Input : arr[] = {5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5}
Output : 5555555550
Input : arr[] = {5, 0}
Output : 0
Since we can choose and permute any number of elements, only the number of 0s and 5s in the array matter. So let’s store the count as c0 and c5 respectively.
The number has to be made a multiple of 90 which is 9*10. Therefore, the number has to be a multiple of both 9 and 10.
The divisibility rules are as follows:
- For a number to be divisible by 10, it should end with 0.
- For a number to be divisible by 9, the sum of digits should be divisible by 9. Since the only non-zero digit allowed to use is 5, the number of times we use 5 has to be a multiple of 9, so that the sum will be a multiple of 45, i.e divisible by 9.
There are 3 possibilities:
- c0=0 . This implies that no number can be made divisible by 10.
- c5=0. This implies that the only number that can be made divisible by 90 is 0.
- If both the above conditions are false. Let’s group the number of 5s into groups of 9. There are going to be floor(c5/9) groups that are completely filled, we can use all of the 5s of all the groups to get the number of 5s a multiple of 9 which also makes the digit sum a multiple of 9. Since increasing the number of zeroes does not affect the divisibility, we can use all the zeroes.
Below is the implementation of the above approach:
C++
// CPP program to find largest number// divisible by 90 that can be made// using 0 and 5#include <bits/stdc++.h>using namespace std;// Function to find largest number// divisible by 90 that can be made// using 0 and 5void printLargestDivisible(int n, int a[]){ // Count of 0s and 5s int i, c0 = 0, c5 = 0; for (i = 0; i < n; i++) { if (a[i] == 0) c0++; else c5++; } // The number of 5s that can be used c5 = floor(c5 / 9) * 9; if (c0 == 0) // The number can't be cout << -1; // made multiple of 10 else if (c5 == 0) // The only multiple of 90 cout << 0; // that can be made is 0 else { for (i = 0; i < c5; i++) cout << 5; for (i = 0; i < c0; i++) cout << 0; }}// Driver Codeint main(){ int a[] = { 5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5 }; int n = sizeof(a) / sizeof(a[0]); printLargestDivisible(n, a); return 0;} |
Java
// Java program to find largest number// divisible by 90 that can be made// using 0 and 5import java.io.*;class GFG {// Function to find largest number// divisible by 90 that can be made// using 0 and 5static void printLargestDivisible(int n, int a[]){ // Count of 0s and 5s int i, c0 = 0, c5 = 0; for (i = 0; i < n; i++) { if (a[i] == 0) c0++; else c5++; } // The number of 5s that can be used c5 = (int)Math.floor(c5 / 9) * 9; if (c0 == 0) // The number can't be System.out.print(-1); // made multiple of 10 else if (c5 == 0) // The only multiple of 90 System.out.println(0); // that can be made is 0 else { for (i = 0; i < c5; i++) System.out.print(5); for (i = 0; i < c0; i++) System.out.print(0); }}// Driver Code public static void main (String[] args) { int a[] = { 5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5 }; int n = a.length; printLargestDivisible(n, a); }}// This code is contributed// by shs |
Python3
# Python3 program to find largest number # divisible by 90 that can be made # using 0 and 5 # from math import every methodsfrom math import *# Function to find largest number # divisible by 90 that can be made # using 0 and 5def printLargestDivisible(n, a) : # Count of 0s and 5s c0, c5 = 0, 0 for i in range(n) : if a[i] == 0 : c0 += 1 else : c5 += 1 # The number of 5s that can be used c5 = floor(c5 / 9) * 9 if c0 == 0 : # The number can't be print(-1,end = "") # made multiple of 10 elif c5 == 0 : # The only multiple of 90 print(0,end = "") # that can be made is 0 else : for i in range(c5) : print(5,end = "") for i in range(c0) : print(0, end = "")# Driver codeif __name__ == "__main__" : a = [ 5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5] n = len(a) # Function calling printLargestDivisible(n, a)# This code is contributed by # ANKITRAI1 |
C#
// C# program to find largest number // divisible by 90 that can be made // using 0 and 5 using System;class GFG {// Function to find largest number // divisible by 90 that can be made // using 0 and 5 public static void printLargestDivisible(int n, int[] a){ // Count of 0s and 5s int i, c0 = 0, c5 = 0; for (i = 0; i < n; i++) { if (a[i] == 0) { c0++; } else { c5++; } } // The number of 5s that can be used c5 = (c5 / 9) * 9; // The number can't be if (c0 == 0) { // made multiple of 10 Console.Write(-1); } // The only multiple of 90 else if (c5 == 0) { // that can be made is 0 Console.WriteLine(0); } else { for (i = 0; i < c5; i++) { Console.Write(5); } for (i = 0; i < c0; i++) { Console.Write(0); } }} // Driver Code public static void Main(string[] args) { int[] a = new int[] {5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5}; int n = a.Length; printLargestDivisible(n, a); }}// This code is contributed by Shrikant13 |
PHP
<?php// PHP program to find largest number // divisible by 90 that can be made // using 0 and 5 // Function to find largest number // divisible by 90 that can be made // using 0 and 5 function printLargestDivisible($n, $a) { // Count of 0s and 5s $i; $c0 = 0; $c5 = 0; for ($i = 0; $i < $n; $i++) { if ($a[$i] == 0) $c0++; else $c5++; } // The number of 5s that can be used $c5 = floor($c5 / 9) * 9; if ($c0 == 0) // The number can't be echo -1; // made multiple of 10 else if ($c5 == 0) // The only multiple of 90 echo 0; // that can be made is 0 else { for ($i = 0; $i < $c5; $i++) echo 5; for ($i = 0; $i < $c0; $i++) echo 0; } } // Driver Code $a = array( 5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5 ); $n = sizeof($a); printLargestDivisible($n, $a); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to find largest number // divisible by 90 that can be made // using 0 and 5 // Function to find largest number // divisible by 90 that can be made // using 0 and 5 function printLargestDivisible(n, a) { // Count of 0s and 5s let i, c0 = 0, c5 = 0; for (i = 0; i < n; i++) { if (a[i] == 0) { c0++; } else { c5++; } } // The number of 5s that can be used c5 = parseInt(c5 / 9, 10) * 9; // The number can't be if (c0 == 0) { // made multiple of 10 document.write(-1); } // The only multiple of 90 else if (c5 == 0) { // that can be made is 0 document.write(0 + "</br>"); } else { for (i = 0; i < c5; i++) { document.write(5); } for (i = 0; i < c0; i++) { document.write(0); } } } let a = [5, 5, 5, 5, 5, 5, 5, 5, 0, 5, 5]; let n = a.length; printLargestDivisible(n, a); </script> |
Output:
5555555550
Time complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(1)
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