# Largest even number that can be formed by any number of swaps

Given an integer N in the form of string, the task is to find the largest even number from the given number when you are allowed to do any number of swaps (swapping the digits of the number). If no even number can be formed then print -1.

Examples:

Input: N = 1324
Output: 4312

Input: N = 135
Output: -1
No even number can be formed using odd digits.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Sort the string in descending order then we will get the largest number possible with the given digit but it may or may not be an even number. In order to make it even (if it not already), an even digit from the number must be swapped with the last digit and in order to maximize the even number the even digit which is to be swapped must the smallest even digit from the number.
Note that the sorting can be done in linear time using frequency array for the digits of the number as the number of distinct elements that are need to be sorted can be at most 10 in the worst case.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `const` `int` `MAX = 10; ` ` `  `// Function to return the maximum ` `// even number that can be formed ` `// with any number of digit swaps ` `string getMaxEven(string str, ``int` `len) ` `{ ` ` `  `    ``// To store the frequencies of ` `    ``// all the digits ` `    ``int` `freq[MAX] = { 0 }; ` ` `  `    ``// To store the minimum even digit ` `    ``// and the minimum overall digit ` `    ``int` `i, minEvenDigit = MAX, minDigit = MAX; ` `    ``for` `(i = 0; i < len; i++) { ` `        ``int` `digit = str[i] - ``'0'``; ` `        ``freq[digit]++; ` ` `  `        ``// If digit is even then update ` `        ``// the minimum even digit ` `        ``if` `(digit % 2 == 0) ` `            ``minEvenDigit = min(digit, minEvenDigit); ` ` `  `        ``// Update the overall minimum digit ` `        ``minDigit = min(digit, minDigit); ` `    ``} ` ` `  `    ``// If there is no even digit then ` `    ``// it is not possible to generate ` `    ``// an even number with swaps ` `    ``if` `(minEvenDigit == MAX) ` `        ``return` `"-1"``; ` ` `  `    ``// Decrease the frequency of the ` `    ``// digits that need to be swapped ` `    ``freq[minEvenDigit]--; ` `    ``freq[minDigit]--; ` ` `  `    ``i = 0; ` `    ``// Take every digit starting from the maximum ` `    ``// in order to maximize the number ` `    ``for` `(``int` `j = MAX - 1; j >= 0; j--) { ` ` `  `        ``// Take current digit number of times ` `        ``// it appeared in the original number ` `        ``for` `(``int` `k = 0; k < freq[j]; k++) ` `            ``str[i++] = (``char``)(j + ``'0'``); ` ` `  `        ``// If current digit equals to the ` `        ``// minimum even digit then one instance of it ` `        ``// needs to be swapped with the minimum overall digit ` `        ``// i.e. append the minimum digit here ` `        ``if` `(j == minEvenDigit) ` `            ``str[i++] = (``char``)(minDigit + ``'0'``); ` `    ``} ` ` `  `    ``// Append once instance of the minimum ` `    ``// even digit in the end to make the number even ` `    ``str[i] = (``char``)(minEvenDigit + ``'0'``); ` ` `  `    ``return` `str; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"1023422"``; ` `    ``int` `len = str.length(); ` ` `  `    ``cout << getMaxEven(str, len); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` `     `  `    ``static` `int` `MAX = ``10``; ` `     `  `    ``// Function to return the maximum ` `    ``// even number that can be formed ` `    ``// with any number of digit swaps ` `    ``static` `String getMaxEven(``char``[] str, ``int` `len) ` `    ``{ ` `     `  `        ``// To store the frequencies of ` `        ``// all the digits ` `        ``int` `[]freq = ``new` `int``[MAX]; ` `     `  `        ``// To store the minimum even digit ` `        ``// and the minimum overall digit ` `        ``int` `i, minEvenDigit = MAX, minDigit = MAX; ` `        ``for` `(i = ``0``; i < len; i++) ` `        ``{ ` `            ``int` `digit = str[i] - ``'0'``; ` `            ``freq[digit]++; ` `     `  `            ``// If digit is even then update ` `            ``// the minimum even digit ` `            ``if` `(digit % ``2` `== ``0``) ` `                ``minEvenDigit = Math.min(digit, minEvenDigit); ` `     `  `            ``// Update the overall minimum digit ` `            ``minDigit = Math.min(digit, minDigit); ` `        ``} ` `     `  `        ``// If there is no even digit then ` `        ``// it is not possible to generate ` `        ``// an even number with swaps ` `        ``if` `(minEvenDigit == MAX) ` `            ``return` `"-1"``; ` `     `  `        ``// Decrease the frequency of the ` `        ``// digits that need to be swapped ` `        ``freq[minEvenDigit]--; ` `        ``freq[minDigit]--; ` `     `  `        ``i = ``0``; ` `         `  `        ``// Take every digit starting from the maximum ` `        ``// in order to maximize the number ` `        ``for` `(``int` `j = MAX - ``1``; j >= ``0``; j--) ` `        ``{ ` `     `  `            ``// Take current digit number of times ` `            ``// it appeared in the original number ` `            ``for` `(``int` `k = ``0``; k < freq[j]; k++) ` `                ``str[i++] = (``char``)(j + ``'0'``); ` `     `  `            ``// If current digit equals to the ` `            ``// minimum even digit then one instance of it ` `            ``// needs to be swapped with the minimum overall digit ` `            ``// i.e. append the minimum digit here ` `            ``if` `(j == minEvenDigit) ` `                ``str[i++] = (``char``)(minDigit + ``'0'``); ` `        ``} ` `     `  `        ``// Append once instance of the minimum ` `        ``// even digit in the end to make the number even ` `        ``str[i-``1``] = (``char``)(minEvenDigit + ``'0'``); ` `     `  `        ``return` `String.valueOf(str); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``char``[] str = ``"1023422"``.toCharArray(); ` `        ``int` `len = str.length; ` `     `  `        ``System.out.println(getMaxEven(str, len)); ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach  ` ` `  `MAX` `=` `10` ` `  `# Function to return the maximum  ` `# even number that can be formed  ` `# with any number of digit swaps  ` `def` `getMaxEven(string, length) : ` `     `  `    ``string ``=` `list``(string) ` `     `  `    ``# To store the frequencies of  ` `    ``# all the digits ` `    ``freq ``=` `[``0``]``*``MAX` `     `  `    ``# To store the minimum even digit ` `    ``# and the minimum overall digit  ` `    ``minEvenDigit ``=` `MAX``; ` `    ``minDigit ``=` `MAX``;  ` `    ``for` `i ``in` `range``(length) : ` `        ``digit ``=` `ord``(string[i]) ``-` `ord``(``'0'``); ` `        ``freq[digit] ``+``=` `1``; ` `         `  `        ``# If digit is even then update  ` `        ``# the minimum even digit ` `        ``if` `(digit ``%` `2` `=``=` `0``) : ` `            ``minEvenDigit ``=` `min``(digit, minEvenDigit); ` `         `  `        ``# Update the overall minimum digit  ` `        ``minDigit ``=` `min``(digit, minDigit); ` `         `  `    ``# If there is no even digit then  ` `    ``# it is not possible to generate  ` `    ``# an even number with swaps  ` `    ``if` `(minEvenDigit ``=``=` `MAX``) : ` `        ``return` `"-1"``;  ` `         `  `    ``# Decrease the frequency of the  ` `    ``# digits that need to be swapped  ` `    ``freq[minEvenDigit] ``-``=` `1``; ` `    ``freq[minDigit] ``-``=` `1``; ` `     `  `    ``i ``=` `0``; ` `     `  `    ``# Take every digit starting from the maximum ` `    ``# in order to maximize the number  ` `    ``for` `j ``in` `range``(``MAX` `-` `1``, ``-``1``, ``-``1``) : ` `         `  `        ``# Take current digit number of times ` `        ``# it appeared in the original number ` `        ``for` `k ``in` `range``(freq[j]) : ` `            ``string[i] ``=` `chr``(j ``+` `ord``(``'0'``)); ` `            ``i ``+``=` `1` `         `  `        ``# If current digit equals to the  ` `        ``# minimum even digit then one instance of it  ` `        ``# needs to be swapped with the minimum overall digit  ` `        ``# i.e. append the minimum digit here ` `        ``if` `(j ``=``=` `minEvenDigit) : ` `            ``string[i] ``=` `chr``(minDigit ``+` `ord``(``'0'``)); ` `            ``i ``+``=` `1` `     `  `    ``# Append once instance of the minimum ` `    ``# even digit in the end to make the number even ` `    ``#string.append(chr(minEvenDigit + ord('0')));  ` `     `  `    ``return` `"".join(string);  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` `    ``string ``=` `"1023422"``;  ` `    ``length ``=` `len``(string);  ` ` `  `    ``print``(getMaxEven(string, length));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `      `  `    ``static` `int` `MAX = 10; ` `      `  `    ``// Function to return the maximum ` `    ``// even number that can be formed ` `    ``// with any number of digit swaps ` `    ``static` `String getMaxEven(``char``[] str, ``int` `len) ` `    ``{ ` `      `  `        ``// To store the frequencies of ` `        ``// all the digits ` `        ``int` `[]freq = ``new` `int``[MAX]; ` `      `  `        ``// To store the minimum even digit ` `        ``// and the minimum overall digit ` `        ``int` `i, minEvenDigit = MAX, minDigit = MAX; ` `        ``for` `(i = 0; i < len; i++) ` `        ``{ ` `            ``int` `digit = str[i] - ``'0'``; ` `            ``freq[digit]++; ` `      `  `            ``// If digit is even then update ` `            ``// the minimum even digit ` `            ``if` `(digit % 2 == 0) ` `                ``minEvenDigit = Math.Min(digit, minEvenDigit); ` `      `  `            ``// Update the overall minimum digit ` `            ``minDigit = Math.Min(digit, minDigit); ` `        ``} ` `      `  `        ``// If there is no even digit then ` `        ``// it is not possible to generate ` `        ``// an even number with swaps ` `        ``if` `(minEvenDigit == MAX) ` `            ``return` `"-1"``; ` `      `  `        ``// Decrease the frequency of the ` `        ``// digits that need to be swapped ` `        ``freq[minEvenDigit]--; ` `        ``freq[minDigit]--; ` `      `  `        ``i = 0; ` `          `  `        ``// Take every digit starting from the maximum ` `        ``// in order to maximize the number ` `        ``for` `(``int` `j = MAX - 1; j >= 0; j--) ` `        ``{ ` `      `  `            ``// Take current digit number of times ` `            ``// it appeared in the original number ` `            ``for` `(``int` `k = 0; k < freq[j]; k++) ` `                ``str[i++] = (``char``)(j + ``'0'``); ` `      `  `            ``// If current digit equals to the ` `            ``// minimum even digit then one instance of it ` `            ``// needs to be swapped with the minimum overall digit ` `            ``// i.e. append the minimum digit here ` `            ``if` `(j == minEvenDigit) ` `                ``str[i++] = (``char``)(minDigit + ``'0'``); ` `        ``} ` `      `  `        ``// Append once instance of the minimum ` `        ``// even digit in the end to make the number even ` `        ``str[i-1] = (``char``)(minEvenDigit + ``'0'``); ` `      `  `        ``return` `String.Join(``""``,str); ` `    ``} ` `      `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``char``[] str = ``"1023422"``.ToCharArray(); ` `        ``int` `len = str.Length; ` `      `  `        ``Console.WriteLine(getMaxEven(str, len)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```4322210
```

Time Complexity: O(n)

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