# Reduce the number to minimum multiple of 4 after removing the digits

Last Updated : 01 Nov, 2023

Given an integer N, the task is to reduce the number to a smallest positive integer X after removing some of the digits (possibly none) such that X is divisible by 4. Print -1 if it cannot be reduced to such multiple.

Examples:

Input: N = 78945666384
Output:
Remove all the digits except a single
occurrence of the digit ‘4’.

Input: N = 17
Output: -1

Approach: Since the resultant number has to be minimized. So, check whether there is any digit in the number which is equal to either ‘4’ or ‘8’ because these are the digits divisible by 4 in the ascending order. If there are no such digits then check all the subsequences of digits of length 2 for any multiple of 4. If there is still no multiple of 4 then the number is not possible because any number with more than 2 digits which is a multiple of 4 will definitely have a subsequence divisible by 4 with digits less 3.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `const` `int` `TEN = 10;`   `// Function to return the minimum number` `// that can be formed after removing` `// the digits which is a multiple of 4` `int` `minNum(string str, ``int` `len)` `{` `    ``int` `res = INT_MAX;`   `    ``// For every digit of the number` `    ``for` `(``int` `i = 0; i < len; i++) {`   `        ``// Check if the current digit` `        ``// is divisible by 4` `        ``if` `(str[i] == ``'4'` `|| str[i] == ``'8'``) {` `            ``res = min(res, str[i] - ``'0'``);` `        ``}` `    ``}`   `    ``for` `(``int` `i = 0; i < len - 1; i++) {` `        ``for` `(``int` `j = i + 1; j < len; j++) {` `            ``int` `num = (str[i] - ``'0'``) * TEN` `                      ``+ (str[j] - ``'0'``);`   `            ``// If any subsequence of two` `            ``// digits is divisible by 4` `            ``if` `(num % 4 == 0) {` `                ``res = min(res, num);` `            ``}` `        ``}` `    ``}`   `    ``return` `((res == INT_MAX) ? -1 : res);` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"17"``;` `    ``int` `len = str.length();`   `    ``cout << minNum(str, len);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.io.*;` `public` `class` `GFG` `{`   `static` `int` `TEN = ``10``;`   `// Function to return the minimum number` `// that can be formed after removing` `// the digits which is a multiple of 4` `static` `int` `minNum(``char``[] str, ``int` `len)` `{` `    ``int` `res = Integer.MAX_VALUE;`   `    ``// For every digit of the number` `    ``for` `(``int` `i = ``0``; i < len; i++)` `    ``{`   `        ``// Check if the current digit` `        ``// is divisible by 4` `        ``if` `(str[i] == ``'4'` `|| str[i] == ``'8'``)` `        ``{` `            ``res = Math.min(res, str[i] - ``'0'``);` `        ``}` `    ``}`   `    ``for` `(``int` `i = ``0``; i < len - ``1``; i++)` `    ``{` `        ``for` `(``int` `j = i + ``1``; j < len; j++) ` `        ``{` `            ``int` `num = (str[i] - ``'0'``) * TEN` `                    ``+ (str[j] - ``'0'``);`   `            ``// If any subsequence of two` `            ``// digits is divisible by 4` `            ``if` `(num % ``4` `== ``0``)` `            ``{` `                ``res = Math.min(res, num);` `            ``}` `        ``}` `    ``}`   `    ``return` `((res == Integer.MAX_VALUE) ? -``1` `: res);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``String str = ``"17"``;` `    ``int` `len = str.length();`   `    ``System.out.print(minNum(str.toCharArray(), len));`   `}` `}`   `// This code is contributed by 29AjayKumar`

## Python 3

 `# Python3 implementation of the approach` `import` `sys` `TEN ``=` `10`   `# Function to return the minimum number` `# that can be formed after removing` `# the digits which is a multiple of 4` `def` `minNum(``str``, len1):` `    ``res ``=` `sys.maxsize`   `    ``# For every digit of the number` `    ``for` `i ``in` `range``(len1):` `        `  `        ``# Check if the current digit` `        ``# is divisible by 4` `        ``if` `(``str``[i] ``=``=` `'4'` `or` `str``[i] ``=``=` `'8'``):` `            ``res ``=` `min``(res, ``ord``(``str``[i]) ``-` `ord``(``'0'``))`   `    ``for` `i ``in` `range``(len1 ``-` `1``):` `        ``for` `j ``in` `range``(i ``+` `1``, len1, ``1``):` `            ``num ``=` `(``ord``(``str``[i]) ``-` `ord``(``'0'``)) ``*` `TEN ``+` `\` `                  ``(``ord``(``str``[j]) ``-` `ord``(``'0'``))`   `            ``# If any subsequence of two` `            ``# digits is divisible by 4` `            ``if` `(num ``%` `4` `=``=` `0``):` `                ``res ``=` `min``(res, num)`   `    ``if` `(res ``=``=` `sys.maxsize):` `        ``return` `-``1` `    ``else``:` `        ``return` `res`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``str` `=` `"17"` `    ``len1 ``=` `len``(``str``)`   `    ``print``(minNum(``str``, len1))` `    `  `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG` `{` `static` `int` `TEN = 10;`   `// Function to return the minimum number` `// that can be formed after removing` `// the digits which is a multiple of 4` `static` `int` `minNum(``char``[] str, ``int` `len)` `{` `    ``int` `res = ``int``.MaxValue;`   `    ``// For every digit of the number` `    ``for` `(``int` `i = 0; i < len; i++)` `    ``{`   `        ``// Check if the current digit` `        ``// is divisible by 4` `        ``if` `(str[i] == ``'4'` `|| str[i] == ``'8'``)` `        ``{` `            ``res = Math.Min(res, str[i] - ``'0'``);` `        ``}` `    ``}`   `    ``for` `(``int` `i = 0; i < len - 1; i++)` `    ``{` `        ``for` `(``int` `j = i + 1; j < len; j++) ` `        ``{` `            ``int` `num = (str[i] - ``'0'``) * TEN` `                    ``+ (str[j] - ``'0'``);`   `            ``// If any subsequence of two` `            ``// digits is divisible by 4` `            ``if` `(num % 4 == 0)` `            ``{` `                ``res = Math.Min(res, num);` `            ``}` `        ``}` `    ``}` `    ``return` `((res == ``int``.MaxValue) ? -1 : res);` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``String str = ``"17"``;` `    ``int` `len = str.Length;`   `    ``Console.Write(minNum(str.ToCharArray(), len));` `}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

```-1
```

Time Complexity: O(n2), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.