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

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. ```

### Approach 1:

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:

1. Sum of digits is already equal to 0 modulo 3. Thus, we don’t have to erase any digits.
2. There exists such a digit that equals sum modulo 3. Then we just have to erase a single digit
3. 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++

 `// CPP program to find the minimum number of``// digits to be removed to make a large number``// divisible by 3.``#include ``using` `namespace` `std;` `// function to count the no of removal of digits``// to make a very large number divisible by 3``int` `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 Code``int` `main()``{``    ``string num = ``"1234"``;``    ``cout << divisible(num);``    ``return` `0;``}`

## Java

 `// 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 3``class` `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

 `# 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 Code``if` `__name__ ``=``=` `'__main__'``:``    ``num ``=` `"1234"``    ``print``(divisible(num))` `# This code is contributed by mits`

## C#

 `// 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 3``class` `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

 ``

## Javascript

 ``

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.

### Approach 2: Without STL Functions:

We can count the sum of digits in the number and check whether it is divisible by 3 or not. If it is divisible by 3, then no digits need to be removed. Otherwise, we need to remove some digits to make it divisible by 3.

This implementation first calculates the sum of digits in the number. Given below is the step-by-step procedure of the above approach:

• If the sum is divisible by 3, then no digits need to be removed, so it returns 0. Otherwise, it tries to remove digits one by one from the right end of the number.
• If a digit can be removed without changing the remainder of the sum modulo 3, then it is removed, and the sum is updated accordingly.
• If removing a single digit is not enough, it tries to remove two digits by choosing the smallest possible digits. If no such combination of digits can make the sum divisible by 3, then the function returns -1 to indicate that it is impossible to make the number divisible by 3 by removing digits.

## C++

 `// CPP program to find the minimum number of``// digits to be removed to make a large number``// divisible by 3.``#include ``using` `namespace` `std;` `int` `divisible(string num) {``    ``int` `n = num.length();``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``sum += num[i] - ``'0'``;``    ``}``    ``if` `(sum % 3 == 0) {``        ``return` `0;``    ``}``    ``int` `cnt1 = 0, cnt2 = 0;``    ``for` `(``int` `i = n-1; i >= 0; i--) {``        ``int` `x = num[i] - ``'0'``;``        ``if` `(x % 3 == sum % 3) {``            ``cnt1++;``            ``sum -= x;``            ``if` `(sum % 3 == 0) {``                ``return` `cnt1;``            ``}``        ``} ``else` `if` `(cnt2 < 2 && x <= sum) {``            ``cnt2++;``            ``sum -= x;``            ``if` `(sum % 3 == 0) {``                ``return` `cnt1 + cnt2;``            ``}``        ``}``    ``}``    ``return` `-1;``}` `    `  `// Driver Code``int` `main()``{``    ``string num = ``"1234"``;``    ``cout << divisible(num);``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {` `    ``public` `static` `int` `divisible(String num) {``        ``int` `n = num.length();``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``sum += num.charAt(i) - ``'0'``;``        ``}``        ``if` `(sum % ``3` `== ``0``) {``            ``return` `0``;``        ``}``        ``int` `cnt1 = ``0``, cnt2 = ``0``;``        ``for` `(``int` `i = n-``1``; i >= ``0``; i--) {``            ``int` `x = num.charAt(i) - ``'0'``;``            ``if` `(x % ``3` `== sum % ``3``) {``                ``cnt1++;``                ``sum -= x;``                ``if` `(sum % ``3` `== ``0``) {``                    ``return` `cnt1;``                ``}``            ``} ``else` `if` `(cnt2 < ``2` `&& x <= sum) {``                ``cnt2++;``                ``sum -= x;``                ``if` `(sum % ``3` `== ``0``) {``                    ``return` `cnt1 + cnt2;``                ``}``            ``}``        ``}``        ``return` `-``1``;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``String num = ``"1234"``;``        ``System.out.println(divisible(num));``    ``}``}`

## Python3

 `def` `divisible(num: ``str``) ``-``> ``int``:``    ``n ``=` `len``(num)``    ``sum` `=` `0``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `int``(num[i])``    ``if` `sum` `%` `3` `=``=` `0``:``        ``return` `0``    ``cnt1, cnt2 ``=` `0``, ``0``    ``for` `i ``in` `range``(n``-``1``, ``-``1``, ``-``1``):``        ``x ``=` `int``(num[i])``        ``if` `x ``%` `3` `=``=` `sum` `%` `3``:``            ``cnt1 ``+``=` `1``            ``sum` `-``=` `x``            ``if` `sum` `%` `3` `=``=` `0``:``                ``return` `cnt1``        ``elif` `cnt2 < ``2` `and` `x <``=` `sum``:``            ``cnt2 ``+``=` `1``            ``sum` `-``=` `x``            ``if` `sum` `%` `3` `=``=` `0``:``                ``return` `cnt1 ``+` `cnt2``    ``return` `-``1` `num ``=` `"1234"``print``(divisible(num))`

## C#

 `using` `System;` `public` `class` `Program``{``  ``public` `static` `int` `Divisible(``string` `num)``  ``{``    ``int` `n = num.Length;``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ``sum += num[i] - ``'0'``;``    ``}``    ``if` `(sum % 3 == 0)``    ``{``      ``return` `0;``    ``}``    ``int` `cnt1 = 0, cnt2 = 0;``    ``for` `(``int` `i = n-1; i >= 0; i--)``    ``{``      ``int` `x = num[i] - ``'0'``;``      ``if` `(x % 3 == sum % 3)``      ``{``        ``cnt1++;``        ``sum -= x;``        ``if` `(sum % 3 == 0)``        ``{``          ``return` `cnt1;``        ``}``      ``}``      ``else` `if` `(cnt2 < 2 && x <= sum)``      ``{``        ``cnt2++;``        ``sum -= x;``        ``if` `(sum % 3 == 0)``        ``{``          ``return` `cnt1 + cnt2;``        ``}``      ``}``    ``}``    ``return` `-1;``  ``}` `  ``public` `static` `void` `Main()``  ``{``    ``string` `num = ``"1234"``;``    ``Console.WriteLine(Divisible(num));``  ``}``}`

## Javascript

 `function` `divisible(num) {``    ``let n = num.length;``    ``let sum = 0;``    ``for` `(let i = 0; i < n; i++) {``        ``sum += parseInt(num[i]);``    ``}``    ``if` `(sum % 3 === 0) {``        ``return` `0;``    ``}``    ``let cnt1 = 0, cnt2 = 0;``    ``for` `(let i = n - 1; i >= 0; i--) {``        ``let x = parseInt(num[i]);``        ``if` `(x % 3 === sum % 3) {``            ``cnt1++;``            ``sum -= x;``            ``if` `(sum % 3 === 0) {``                ``return` `cnt1;``            ``}``        ``} ``else` `if` `(cnt2 < 2 && x <= sum) {``            ``cnt2++;``            ``sum -= x;``            ``if` `(sum % 3 === 0) {``                ``return` `cnt1 + cnt2;``            ``}``        ``}``    ``}``    ``return` `-1;``}` `let num = ``"1234"``;``console.log(divisible(num));`

Output

`1`

Time Complexity: O(N), as we use a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.