Find the largest multiple of 3 from array of digits | Set 2 (In O(n) time and O(1) space)
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++
// 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// countsvoid 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 ind2bool 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 codeint main(){ int arr[] = {4, 4, 1, 1, 1, 3 } ; int n = sizeof(arr)/sizeof(arr[0]); largest3Multiple(arr, n); return 0;} |
Java
// Java program to find the largest number// that can be mode from elements of the// array and is divisible by 3import 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
# Python3 program to find the largest number# that can be mode from elements of the# array and is divisible by 3 # Number of digitsMAX_SIZE = 10 # function to sort array of digits using# countsdef 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 ind2def 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 codeif __name__ == "__main__": arr = [4, 4, 1, 1, 1, 3] n = len(arr) largest3Multiple(arr, n) # This code is contributed by# sanjeev2552 |
C#
// C# program to find the largest number// that can be mode from elements of the// array and is divisible by 3using 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 |
Javascript
<script>// JavaScript program to find the largest number// that can be mode from elements of the// array and is divisible by 3 // Number of digitsconst MAX_SIZE = 10; // function to sort array of digits using// countsfunction 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 ind2function 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)
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