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 Practice

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 needed 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 swapsstring 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;    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);    }     // 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]--;     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');    }     // Append once instance of the minimum    // even digit in the end to make the number even    str[i] = (char)(minEvenDigit + '0');     return str;} // Driver codeint main(){    string str = "1023422";    int len = str.length();     // Function call    cout << getMaxEven(str, len);     return 0;}

Java

 // Java implementation of the approachimport java.io.*;public 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(String str, int len)    {      //To store the max even number        String maxEven="";        // To store the frequencies of        // all the digits        int[] freq = new int[MAX];         // To store the minimum even digit        int i, minEvenDigit = MAX;        for (i = 0; i < len; i++) {            int digit = str.charAt(i) - '0';            freq[digit]++;             // If digit is even then update            // the minimum even digit            if (digit % 2 == 0)                minEvenDigit                    = Math.min(digit, minEvenDigit);         }         // 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        // minEvenDigit        freq[minEvenDigit]--;                  i = MAX-1;         // Take every digit starting from the maximum        // in order to maximize the number       while(i>=0)       {           // Take current digit number of times            // it appeared in the original number           if(freq[i]>0)           {             maxEven= maxEven+i;             freq[i]--;           }else             i--;                     }         // Append the minimum even digit        // in the end to make the number even         maxEven= maxEven+minEvenDigit;         return maxEven;    }     // Driver code    public static void main(String[] args)    {        String str = "1023422";        int len = str.length();         // Function call        System.out.println(getMaxEven(str, len));    }}

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[-1] = chr(minEvenDigit + ord('0'))     return "".join(string)  # Driver codeif __name__ == "__main__":    string = "1023422"    length = len(string)     # Function call    print(getMaxEven(string, length)) # This code is contributed by AnkitRai01

C#

 // C# implementation of the approachusing 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;         // Function call        Console.WriteLine(getMaxEven(str, len));    }} // This code has been contributed by 29AjayKumar

Javascript



Output:
4322210

Time Complexity: O(n + MAX)
Auxiliary Space: O(MAX)

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