Given an array of digits (contain elements from 0 to 9). Find the largest number that can be made from some or all digits of array and is divisible by 3. The same element may appear multiple times in the array, but each element in the array may only be used once.
Examples:
Input : arr[] = {5, 4, 3, 1, 1} Output : 4311 Input : Arr[] = {5, 5, 5, 7} Output : 555
Asked In : Google Interview
We have discussed a queue based solution. Both solutions (discussed in previous and this posts) are based on the fact that a number is divisible by 3 if and only if sum of digits of the number is divisible by 3.
For example, let us consider 555, it is divisible by 3 because sum of digits is 5 + 5 + 5 = 15, which is divisible by 3. If a sum of digits is not divisible by 3 then the remainder should be either 1 or 2.
If we get remainder either ‘1’ or ‘2’, we have to remove maximum two digits to make a number that is divisible by 3:
- If remainder is ‘1’ : We have to remove single digit that have remainder ‘1’ or we have to remove two digit that have remainder ‘2’ ( 2 + 2 => 4 % 3 => ‘1’)
- If remainder is ‘2’ : .We have to remove single digit that have remainder ‘2’ or we have to remove two digit that have remainder ‘1’ ( 1 + 1 => 2 % 3 => 2 ).
Examples :
Input : arr[] = 5, 5, 5, 7 Sum of digits = 5 + 5 + 5 + 7 = 22 Remainder = 22 % 3 = 1 We remove smallest single digit that has remainder '1'. We remove 7 % 3 = 1 So largest number divisible by 3 is : 555 Let's take an another example : Input : arr[] = 4 , 4 , 1 , 1 , 1 , 3 Sum of digits = 4 + 4 + 1 + 1 + 1 + 3 = 14 Reminder = 14 % 3 = 2 We have to remove the smallest digit that has remainder ' 2 ' or two digits that have remainder '1'. Here there is no digit with reminder '2', so we have to remove two smallest digits that have remainder '1'. The digits are : 1, 1. So largest number divisible by 3 is 4 4 3 1
Below are implementation of above idea.
// C++ program to find the largest number // that can be mode from elements of the // array and is divisible by 3 #include<bits/stdc++.h> using namespace std;
// Number of digits #define MAX_SIZE 10 // function to sort array of digits using // counts void sortArrayUsingCounts( int arr[], int n)
{ // Store count of all elements
int count[MAX_SIZE] = {0};
for ( int i = 0; i < n; i++)
count[arr[i]]++;
// Store
int index = 0;
for ( int i = 0; i < MAX_SIZE; i++)
while (count[i] > 0)
arr[index++] = i, count[i]--;
} // Remove elements from arr[] at indexes ind1 and ind2 bool removeAndPrintResult( int arr[], int n, int ind1,
int ind2 = -1)
{ for ( int i = n-1; i >=0; i--)
if (i != ind1 && i != ind2)
cout << arr[i] ;
} // Returns largest multiple of 3 that can be formed // using arr[] elements. bool largest3Multiple( int arr[], int n)
{ // Sum of all array element
int sum = accumulate(arr, arr+n, 0);
// Sort array element in increasing order
sortArrayUsingCounts(arr, n);
// Sum is divisible by 3 , no need to
// delete an element
if (sum%3 == 0)
{
removeAndPrintResult(arr,n,-1,-1);
return true ;
}
// Find reminder
int remainder = sum % 3;
// If remainder is '1', we have to delete either
// one element of remainder '1' or two elements
// of remainder '2'
if (remainder == 1)
{
int rem_2[2];
rem_2[0] = -1, rem_2[1] = -1;
// Traverse array elements
for ( int i = 0 ; i < n ; i++)
{
// Store first element of remainder '1'
if (arr[i]%3 == 1)
{
removeAndPrintResult(arr, n, i);
return true ;
}
if (arr[i]%3 == 2)
{
// If this is first occurrence of remainder 2
if (rem_2[0] == -1)
rem_2[0] = i;
// If second occurrence
else if (rem_2[1] == -1)
rem_2[1] = i;
}
}
if (rem_2[0] != -1 && rem_2[1] != -1)
{
removeAndPrintResult(arr, n, rem_2[0], rem_2[1]);
return true ;
}
}
// If remainder is '2', we have to delete either
// one element of remainder '2' or two elements
// of remainder '1'
else if (remainder == 2)
{
int rem_1[2];
rem_1[0] = -1, rem_1[1] = -1;
// traverse array elements
for ( int i = 0; i < n; i++)
{
// store first element of remainder '2'
if (arr[i]%3 == 2)
{
removeAndPrintResult(arr, n, i);
return true ;
}
if (arr[i]%3 == 1)
{
// If this is first occurrence of remainder 1
if (rem_1[0] == -1)
rem_1[0] = i;
// If second occurrence
else if (rem_1[1] == -1)
rem_1[1] = i;
}
}
if (rem_1[0] != -1 && rem_1[1] != -1)
{
removeAndPrintResult(arr, n, rem_1[0], rem_1[1]);
return true ;
}
}
cout << "Not possible" ;
return false ;
} // Driver code int main()
{ int arr[] = {4, 4, 1, 1, 1, 3 } ;
int n = sizeof (arr)/ sizeof (arr[0]);
largest3Multiple(arr, n);
return 0;
} |
// Java program to find the largest number // that can be mode from elements of the // array and is divisible by 3 import java.util.*;
class GFG
{ // Number of digits
static int MAX_SIZE = 10 ;
// function to sort array of digits using
// counts
static void sortArrayUsingCounts( int arr[],
int n)
{
// Store count of all elements
int [] count = new int [MAX_SIZE];
for ( int i = 0 ; i < n; i++)
{
count[arr[i]]++;
}
// Store
int index = 0 ;
for ( int i = 0 ; i < MAX_SIZE; i++)
{
while (count[i] > 0 )
{
arr[index++] = i;
count[i]--;
}
}
}
// Remove elements from arr[]
// at indexes ind1 and ind2
static void removeAndPrintResult( int arr[], int n,
int ind1, int ind2)
{
for ( int i = n - 1 ; i >= 0 ; i--)
{
if (i != ind1 && i != ind2)
{
System.out.print(arr[i]);
}
}
}
// Returns largest multiple of 3
// that can be formed using
// arr[] elements.
static boolean largest3Multiple( int arr[],
int n)
{
// Sum of all array element
int sum = accumulate(arr, 0 , n);
// Sort array element in increasing order
sortArrayUsingCounts(arr, n);
// If sum is divisible by 3,
// no need to delete an element
if (sum % 3 == 0 )
{
removeAndPrintResult(arr, n, - 1 , - 1 );
return true ;
}
// Find reminder
int remainder = sum % 3 ;
// If remainder is '1', we have to
// delete either one element of
// remainder '1' or two elements of
// remainder '2'
if (remainder == 1 )
{
int [] rem_2 = new int [ 2 ];
rem_2[ 0 ] = - 1 ;
rem_2[ 1 ] = - 1 ;
// Traverse array elements
for ( int i = 0 ; i < n; i++)
{
// Store first element of remainder '1'
if (arr[i] % 3 == 1 )
{
removeAndPrintResult(arr, n, i, - 1 );
return true ;
}
if (arr[i] % 3 == 2 )
{
// If this is first occurrence
// of remainder 2
if (rem_2[ 0 ] == - 1 )
{
rem_2[ 0 ] = i;
}
// If second occurrence
else if (rem_2[ 1 ] == - 1 )
{
rem_2[ 1 ] = i;
}
}
}
if (rem_2[ 0 ] != - 1 &&
rem_2[ 1 ] != - 1 )
{
removeAndPrintResult(arr, n, rem_2[ 0 ],
rem_2[ 1 ]);
return true ;
}
}
// If remainder is '2', we have to
// delete either one element of
// remainder '2' or two elements of
// remainder '1'
else if (remainder == 2 )
{
int [] rem_1 = new int [ 2 ];
rem_1[ 0 ] = - 1 ;
rem_1[ 1 ] = - 1 ;
// traverse array elements
for ( int i = 0 ; i < n; i++)
{
// store first element of remainder '2'
if (arr[i] % 3 == 2 )
{
removeAndPrintResult(arr, n, i, - 1 );
return true ;
}
if (arr[i] % 3 == 1 )
{
// If this is first occurrence
// of remainder 1
if (rem_1[ 0 ] == - 1 )
{
rem_1[ 0 ] = i;
}
// If second occurrence
else if (rem_1[ 1 ] == - 1 )
{
rem_1[ 1 ] = i;
}
}
}
if (rem_1[ 0 ] != - 1 &&
rem_1[ 1 ] != - 1 )
{
removeAndPrintResult(arr, n, rem_1[ 0 ],
rem_1[ 1 ]);
return true ;
}
}
System.out.print( "Not possible" );
return false ;
}
static int accumulate( int [] arr,
int start,
int end)
{
int sum = 0 ;
for ( int i = 0 ; i < arr.length; i++)
{
sum += arr[i];
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 4 , 4 , 1 , 1 , 1 , 3 };
int n = arr.length;
largest3Multiple(arr, n);
}
} |
# Python3 program to find the largest number # that can be mode from elements of the # array and is divisible by 3 # Number of digits MAX_SIZE = 10
# function to sort array of digits using # counts def sortArrayUsingCounts(arr, n):
# Store count of all elements
count = [ 0 ] * MAX_SIZE
for i in range (n):
count[arr[i]] + = 1
# Store
index = 0
for i in range (MAX_SIZE):
while count[i] > 0 :
arr[index] = i
index + = 1
count[i] - = 1
# Remove elements from arr[] at indexes ind1 and ind2 def removeAndPrintResult(arr, n, ind1, ind2 = - 1 ):
for i in range (n - 1 , - 1 , - 1 ):
if i ! = ind1 and i ! = ind2:
print (arr[i], end = "")
# Returns largest multiple of 3 that can be formed # using arr[] elements. def largest3Multiple(arr, n):
# Sum of all array element
s = sum (arr)
# Sort array element in increasing order
sortArrayUsingCounts(arr, n)
# Sum is divisible by 3, no need to
# delete an element
if s % 3 = = 0 :
removeAndPrintResult(arr, n, - 1 )
return True
# Find reminder
remainder = s % 3
# If remainder is '1', we have to delete either
# one element of remainder '1' or two elements
# of remainder '2'
if remainder = = 1 :
rem_2 = [ 0 ] * 2
rem_2[ 0 ] = - 1 ; rem_2[ 1 ] = - 1
# Traverse array elements
for i in range (n):
# Store first element of remainder '1'
if arr[i] % 3 = = 1 :
removeAndPrintResult(arr, n, i)
return True
if arr[i] % 3 = = 2 :
# If this is first occurrence of remainder 2
if rem_2[ 0 ] = = - 1 :
rem_2[ 0 ] = i
# If second occurrence
elif rem_2[ 1 ] = = - 1 :
rem_2[ 1 ] = i
if rem_2[ 0 ] ! = - 1 and rem_2[ 1 ] ! = - 1 :
removeAndPrintResult(arr, n, rem_2[ 0 ], rem_2[ 1 ])
return True
# If remainder is '2', we have to delete either
# one element of remainder '2' or two elements
# of remainder '1'
elif remainder = = 2 :
rem_1 = [ 0 ] * 2
rem_1[ 0 ] = - 1 ; rem_1[ 1 ] = - 1
# traverse array elements
for i in range (n):
# store first element of remainder '2'
if arr[i] % 3 = = 2 :
removeAndPrintResult(arr, n, i)
return True
if arr[i] % 3 = = 1 :
# If this is first occurrence of remainder 1
if rem_1[ 0 ] = = - 1 :
rem_1[ 0 ] = i
# If second occurrence
elif rem_1[ 1 ] = = - 1 :
rem_1[ 1 ] = i
if rem_1[ 0 ] ! = - 1 and rem_1[ 1 ] ! = - 1 :
removeAndPrintResult(arr, n, rem_1[ 0 ], rem_1[ 1 ])
return True
print ( "Not possible" )
return False
# Driver code if __name__ = = "__main__" :
arr = [ 4 , 4 , 1 , 1 , 1 , 3 ]
n = len (arr)
largest3Multiple(arr, n)
# This code is contributed by # sanjeev2552 |
// C# program to find the largest number // that can be mode from elements of the // array and is divisible by 3 using System;
class GFG
{ // Number of digits
static int MAX_SIZE = 10;
// function to sort array of digits using
// counts
static void sortArrayUsingCounts( int []arr,
int n)
{
// Store count of all elements
int [] count = new int [MAX_SIZE];
for ( int i = 0; i < n; i++)
{
count[arr[i]]++;
}
// Store
int index = 0;
for ( int i = 0; i < MAX_SIZE; i++)
{
while (count[i] > 0)
{
arr[index++] = i;
count[i]--;
}
}
}
// Remove elements from arr[]
// at indexes ind1 and ind2
static void removeAndPrintResult( int []arr, int n,
int ind1, int ind2)
{
for ( int i = n - 1; i >= 0; i--)
{
if (i != ind1 && i != ind2)
{
Console.Write(arr[i]);
}
}
}
// Returns largest multiple of 3
// that can be formed using
// arr[] elements.
static Boolean largest3Multiple( int []arr,
int n)
{
// Sum of all array element
int sum = accumulate(arr, 0, n);
// Sort array element in increasing order
sortArrayUsingCounts(arr, n);
// If sum is divisible by 3,
// no need to delete an element
if (sum % 3 == 0)
{
removeAndPrintResult(arr, n, -1, -1);
return true ;
}
// Find reminder
int remainder = sum % 3;
// If remainder is '1', we have to
// delete either one element of
// remainder '1' or two elements of
// remainder '2'
if (remainder == 1)
{
int [] rem_2 = new int [2];
rem_2[0] = -1;
rem_2[1] = -1;
// Traverse array elements
for ( int i = 0; i < n; i++)
{
// Store first element of remainder '1'
if (arr[i] % 3 == 1)
{
removeAndPrintResult(arr, n, i, -1);
return true ;
}
if (arr[i] % 3 == 2)
{
// If this is first occurrence
// of remainder 2
if (rem_2[0] == -1)
{
rem_2[0] = i;
}
// If second occurrence
else if (rem_2[1] == -1)
{
rem_2[1] = i;
}
}
}
if (rem_2[0] != -1 &&
rem_2[1] != -1)
{
removeAndPrintResult(arr, n, rem_2[0],
rem_2[1]);
return true ;
}
}
// If remainder is '2', we have to
// delete either one element of
// remainder '2' or two elements of
// remainder '1'
else if (remainder == 2)
{
int [] rem_1 = new int [2];
rem_1[0] = -1;
rem_1[1] = -1;
// traverse array elements
for ( int i = 0; i < n; i++)
{
// store first element of remainder '2'
if (arr[i] % 3 == 2)
{
removeAndPrintResult(arr, n, i, -1);
return true ;
}
if (arr[i] % 3 == 1)
{
// If this is first occurrence
// of remainder 1
if (rem_1[0] == -1)
{
rem_1[0] = i;
}
// If second occurrence
else if (rem_1[1] == -1)
{
rem_1[1] = i;
}
}
}
if (rem_1[0] != -1 &&
rem_1[1] != -1)
{
removeAndPrintResult(arr, n, rem_1[0],
rem_1[1]);
return true ;
}
}
Console.Write( "Not possible" );
return false ;
}
static int accumulate( int [] arr,
int start,
int end)
{
int sum = 0;
for ( int i = 0; i < arr.Length; i++)
{
sum += arr[i];
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {4, 4, 1, 1, 1, 3};
int n = arr.Length;
largest3Multiple(arr, n);
}
} // This code is contributed by 29AjayKumar |
<script> // JavaScript program to find the largest number // that can be mode from elements of the // array and is divisible by 3 // Number of digits const MAX_SIZE = 10; // function to sort array of digits using // counts function sortArrayUsingCounts(arr, n)
{ // Store count of all elements
let count = new Uint8Array(MAX_SIZE);
for (let i = 0; i < n; i++)
count[arr[i]]++;
// Store
let index = 0;
for (let i = 0; i < MAX_SIZE; i++)
while (count[i] > 0)
arr[index++] = i, count[i]--;
} // Remove elements from arr[] at indexes ind1 and ind2 function removeAndPrintResult(arr, n, ind1, ind2 = -1)
{ for (let i = n-1; i >=0; i--)
if (i != ind1 && i != ind2)
document.write(arr[i]) ;
} // Returns largest multiple of 3 that can be formed // using arr[] elements. function largest3Multiple(arr, n)
{ // Sum of all array element
let sum = arr.reduce((a, b) => a + b, 0);
// Sort array element in increasing order
sortArrayUsingCounts(arr, n);
// Sum is divisible by 3 , no need to
// delete an element
if (sum%3 == 0)
{
removeAndPrintResult(arr, n, -1);
return true ;
}
// Find reminder
let remainder = sum % 3;
// If remainder is '1', we have to delete either
// one element of remainder '1' or two elements
// of remainder '2'
if (remainder == 1)
{
let rem_2 = new Array(2);
rem_2[0] = -1, rem_2[1] = -1;
// Traverse array elements
for (let i = 0 ; i < n ; i++)
{
// Store first element of remainder '1'
if (arr[i]%3 == 1)
{
removeAndPrintResult(arr, n, i);
return true ;
}
if (arr[i]%3 == 2)
{
// If this is first occurrence of remainder 2
if (rem_2[0] == -1)
rem_2[0] = i;
// If second occurrence
else if (rem_2[1] == -1)
rem_2[1] = i;
}
}
if (rem_2[0] != -1 && rem_2[1] != -1)
{
removeAndPrintResult(arr, n, rem_2[0], rem_2[1]);
return true ;
}
}
// If remainder is '2', we have to delete either
// one element of remainder '2' or two elements
// of remainder '1'
else if (remainder == 2)
{
let rem_1 = new Array(2);
rem_1[0] = -1, rem_1[1] = -1;
// traverse array elements
for (let i = 0; i < n; i++)
{
// store first element of remainder '2'
if (arr[i]%3 == 2)
{
removeAndPrintResult(arr, n, i);
return true ;
}
if (arr[i]%3 == 1)
{
// If this is first occurrence of remainder 1
if (rem_1[0] == -1)
rem_1[0] = i;
// If second occurrence
else if (rem_1[1] == -1)
rem_1[1] = i;
}
}
if (rem_1[0] != -1 && rem_1[1] != -1)
{
removeAndPrintResult(arr, n, rem_1[0], rem_1[1]);
return true ;
}
}
document.write( "Not possible" );
return false ;
} // Driver code let arr = [4 , 4 , 1 , 1 , 1 , 3];
let n = arr.length;
largest3Multiple(arr, n);
// This code is contributed by Surbhi Tyagi. </script> |
Output:
4431
Time Complexity : O(n)