Number of digits to be removed to make a number divisible by 3

08 Jun, 2022
Given a very large number num (1 <= num <= 10^1000), print the number of digits that needs to be removed to make the number exactly divisible by 3. If it is not possible then print -1.
Examples : 
Input: num = "1234"
Output: 1
Explanation: we need to remove one 
digit that is 1 or 4, to make the
number divisible by 3.on 

Input: num = "11"
Output: -1
Explanation: It is not possible to 
remove any digits and make it divisible
by 3. 
The idea is based on the fact that a number is multiple of 3 if and only if sum of its digits is multiple of 3 (See this for details).One important observation used here is that the answer is at most 2 if an answer exists. So here are the only options for the function: 
Sum of digits is already equal to 0 modulo 3. Thus, we don’t have to erase any digits.There exists such a digit that equals sum modulo 3. Then we just have to erase a single digitAll the digits are neither divisible by 3 nor equal to sum modulo 3. So two of such digits will sum up to number, which equals sum modulo 3, (2+2) mod 3=1, (1+1) mod 3=2
Sum of digits is already equal to 0 modulo 3. Thus, we don’t have to erase any digits.
There exists such a digit that equals sum modulo 3. Then we just have to erase a single digit
All the digits are neither divisible by 3 nor equal to sum modulo 3. So two of such digits will sum up to number, which equals sum modulo 3, (2+2) mod 3=1, (1+1) mod 3=2
C++
Java
Python3
C#
PHP
Javascript
// CPP program to find the minimum number of// digits to be removed to make a large number// divisible by 3.#include <bits/stdc++.h>using namespace std; // function to count the no of removal of digits// to make a very large number divisible by 3int divisible(string num){    int n = num.length();     // add up all the digits of num    int sum = accumulate(begin(num),                         end(num), 0) - '0' * 1;     // if num is already is divisible by 3    // then no digits are to be removed    if (sum % 3 == 0)        return 0;     // if there is single digit, then it is    // not possible to remove one digit.    if (n == 1)        return -1;     // traverse through the number and find out    // if any number on removal makes the sum    // divisible by 3    for (int i = 0; i < n; i++)        if (sum % 3 == (num[i] - '0') % 3)            return 1;     // if there are two numbers then it is    // not possible to remove two digits.    if (n == 2)        return -1;     // Otherwise we can always make a number    // multiple of 2 by removing 2 digits.    return 2;} // Driver Codeint main(){    string num = "1234";    cout << divisible(num);    return 0;}
// Java program to find the// minimum number of digits// to be removed to make a// large number divisible by 3.import java.io.*; // function to count the no// of removal of digits// to make a very large// number divisible by 3class GFG {    static int divisible(String num)    {        int n = num.length();         // add up all the        // digits of num        int sum = 0;        for (int i = 0; i < n; i++)            sum += (int)(num.charAt(i));         // if num is already is        // divisible by 3 then        // no digits are to be        // removed        if (sum % 3 == 0)            return 0;         // if there is single digit,        // then it is not possible        // to remove one digit.        if (n == 1)            return -1;         // traverse through the number        // and find out if any number        // on removal makes the sum        // divisible by 3        for (int i = 0; i < n; i++)            if (sum % 3 == (num.charAt(i) - '0') % 3)                return 1;         // if there are two numbers        // then it is not possible        // to remove two digits.        if (n == 2)            return -1;         // Otherwise we can always        // make a number multiple        // of 2 by removing 2 digits.        return 2;    }     // Driver Code    public static void main(String[] args)    {         String num = "1234";        System.out.println(divisible(num));    }} // This code is contributed by Raj
# Python3 program to find the# minimum number of digits to# be removed to make a large# number divisible by 3. # function to count the# no of removal of digits# to make a very large# number divisible by 3  def divisible(num):    n = len(num)     # add up all the digits of num    sum_ = 0    for i in range(n):        sum_ += int(num[i])     # if num is already is    # divisible by 3 then no    # digits are to be removed    if (sum_ % 3 == 0):        return 0     # if there is single digit,    # then it is not possible    # to remove one digit.    if (n == 1):        return -1     # traverse through the number    # and find out if any number    # on removal makes the sum    # divisible by 3    for i in range(n):        if (sum_ % 3 == int(num[i]) % 3):            return 1     # if there are two numbers    # then it is not possible    # to remove two digits.    if (n == 2):        return -1     # Otherwise we can always    # make a number multiple of    # 2 by removing 2 digits.    return 2  # Driver Codeif __name__ == '__main__':    num = "1234"    print(divisible(num)) # This code is contributed by mits
// C# program to find the// minimum number of digits// to be removed to make a// large number divisible by 3.using System; // function to count the no// of removal of digits// to make a very large// number divisible by 3class GFG {    static int divisible(String num)    {        int n = num.Length;         // add up all the        // digits of num        int sum = 0;        for (int i = 0; i < n; i++)            sum += (int)(num[i]);         // if num is already is        // divisible by 3 then        // no digits are to be        // removed        if (sum % 3 == 0)            return 0;         // if there is single digit,        // then it is not possible        // to remove one digit.        if (n == 1)            return -1;         // traverse through the number        // and find out if any number        // on removal makes the sum        // divisible by 3        for (int i = 0; i < n; i++)            if (sum % 3 == (num[i] - '0') % 3)                return 1;         // if there are two numbers        // then it is not possible        // to remove two digits.        if (n == 2)            return -1;         // Otherwise we can always        // make a number multiple        // of 2 by removing 2 digits.        return 2;    }     // Driver Code    public static void Main()    {        string num = "1234";        Console.WriteLine(divisible(num));    }} // This code is contributed by mits
<?php// PHP program to find the// minimum number of digits to// be removed to make a large// number divisible by 3. // function to count the// no of removal of digits// to make a very large// number divisible by 3 function divisible($num){    $n = strlen($num);     // add up all the digits of num    $sum = ($num); ($num); 0 - '0';     // if num is already is    // divisible by 3 then no    // digits are to be removed    if ($sum % 3 == 0)        return 0;     // if there is single digit,    // then it is not possible    // to remove one digit.    if ($n == 1)        return -1;     // traverse through the number    // and find out if any number    // on removal makes the sum    // divisible by 3    for ($i = 0; $i < $n; $i++)        if ($sum % 3 == ($num[$i] - '0') % 3)            return 1;             // if there are two numbers    // then it is not possible    // to remove two digits.    if ($n == 2)        return -1;     // Otherwise we can always    // make a number multiple of    // 2 by removing 2 digits.    return 2;} // Driver Code$num = "1234";echo divisible($num); // This code is contributed by ajit?>
<script> // JavaScript program to find the minimum number of// digits to be removed to make a large number// divisible by 3. // function to count the no of removal of digits// to make a very large number divisible by 3function divisible(num){    let n = num.length;     // add up all the digits of num    let sum = 0;    for (let i = 0; i < n; i++)            sum += (num.charAt(i));     // if num is already is divisible by 3    // then no digits are to be removed    if (sum % 3 == 0)        return 0;     // if there is single digit, then it is    // not possible to remove one digit.    if (n == 1)        return -1;     // traverse through the number and find out    // if any number on removal makes the sum    // divisible by 3    for (let i = 0; i < n; i++)        if (sum % 3 == (num.charAt(i) - '0') % 3)            return 1;     // if there are two numbers then it is    // not possible to remove two digits.    if (n == 2)        return -1;     // Otherwise we can always make a number    // multiple of 2 by removing 2 digits.    return 2;} // Driver Code     let num = "1234";    document.write(divisible(num));  // This code is contributed by Surbhi Tyagi. </script>
Output : 
1
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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Mithun Kumar
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