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Smallest even digits number not less than N

Last Updated : 16 Dec, 2022
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Given a number N, we need to write a program to find the smallest number not less than N, which has all digits even. 
Examples: 

Input: N = 1345  
Output: 2000
Explanation: 2000 is the smallest number not less than N, whose all digits are even. 

Input : N = 2397
Output : 2400 
Explanation: 2400 is the smallest number not less than N, whose all digits are even.


Naive approach: A naive approach is to keep iterating from N until we find a number with all digits even. 

Below is the implementation of above approach: 

C++ 
// CPP program to print the smallest
// integer not less than N with all even digits
#include <bits/stdc++.h>
using namespace std;

// function to check if all digits
// are even of a given number
int check_digits(int n)
{
    // iterate for all digits
    while (n) {
        if ((n % 10) % 2) // if digit is odd
            return 0;

        n /= 10;
    }

    // all digits are even
    return 1;
}

// function to return the smallest number
// with all digits even
int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits even
    for (int i = n;; i++)
        if (check_digits(i))
            return i;
}

// Driver Code
int main()
{
    int N = 2397;
    cout << smallest_number(N);
    return 0;
}
Java 
// Java program to print the smallest
// integer not less than N with all 
// even digits
import java.io.*;
public class GFG {
    
    // function to check if all digits
    // are even of a given number
    static int check_digits(int n)
    {
        
        // iterate for all digits
        while (n != 0) {
            
            // if digit is odd
            if ((n % 10) % 2 != 0)
                return 0;

            n /= 10;
        }

        // all digits are even
        return 1;
    }

    // function to return the smallest
    // number with all digits even
    static int smallest_number(int n)
    {
        
        // iterate till we find a
        // number with all digits even
        for (int i = n; ; i++)
            if (check_digits(i) != 0)
                return i;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 2397;
        
        System.out.println(smallest_number(N));
    }
}

// This code is contributed by
// Smitha Dinesh Semwal
Python3 
# Python3 program to print the smallest
# integer not less than N with 
# all even digits

# function to check if all digits
# are even of a given number
def check_digits(n) :
    
    # iterate for all digits
    while (n) : 
        
        # if digit is odd
        if ((n % 10) % 2) :
            return 0

        n = int(n / 10)
        
    # all digits are even
    return 1

# function to return the
# smallest number with 
# all digits even
def smallest_number(n) :
    
    # iterate till we find a
    # number with all digits even
    for i in range(n, 2401) :
        if (check_digits(i) == 1) :
            return (i)

# Driver Code
N = 2397
print (str(smallest_number(N)))

# This code is contributed by 
# Manish Shaw (manishshaw1)
C# 
// C# program to print the smallest
// integer not less than N with all 
// even digits
using System;
class GFG {
    
    // function to check if all digits
    // are even of a given number
    static int check_digits(int n)
    {
        
        // iterate for all digits
        while (n != 0) {
            
            // if digit is odd
            if ((n % 10) % 2 != 0)
                return 0;

            n /= 10;
        }

        // all digits are even
        return 1;
    }

    // function to return the smallest
    // number with all digits even
    static int smallest_number(int n)
    {
        
        // iterate till we find a
        // number with all digits even
        for (int i = n; ; i++)
            if (check_digits(i) != 0)
                return i;
    }

    // Driver Code
    public static void Main()
    {
        int N = 2397;
        
    Console.WriteLine(smallest_number(N));
    }
}

// This code is contributed by anuj_67.
PHP 
<?php
// PHP program to print the smallest
// integer not less than N with 
// all even digits

// function to check if all digits
// are even of a given number
function check_digits($n)
{
    
    // iterate for all digits
    while ($n) 
    {
        
        // if digit is odd
        if (($n % 10) % 2) 
            return 0;

        $n /= 10;
    }

    // all digits are even
    return 1;
}

// function to return the
// smallest number with 
// all digits even
function smallest_number( $n)
{
    
    // iterate till we find a
    // number with all digits even
    for ($i = $n; ; $i++)
        if (check_digits($i))
            return $i;
}

    // Driver Code
    $N = 2397;
    echo smallest_number($N);
    
// This code is contributed by m_kit
?>
JavaScript 
<script>

// Javascript program to print the smallest
// integer not less than N with all 
// even digits

    
// function to check if all digits
// are even of a given number
function check_digits(n)
{
    
    // iterate for all digits
    while (n != 0) {
        
        // if digit is odd
        if ((n % 10) % 2 != 0)
            return 0;

        n = parseInt(n/10);
    }

    // all digits are even
    return 1;
}

// function to return the smallest
// number with all digits even
function smallest_number(n)
{
    
    // iterate till we find a
    // number with all digits even
    for (i = n; ; i++)
        if (check_digits(i) != 0)
            return i;
}

// Driver Code
var N = 2397;

document.write(smallest_number(N));

// This code is contributed by 29AjayKumar 

</script>

Output
2400

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: We can find the number by increasing the first odd digit in N by one and replacing all digits to the right of that odd digit with the smallest even digit (i.e. 0). If there are no odd digits in N, then N is the smallest number itself. For example, consider N = 213. Increment first odd digit in N i.e., 1 to 2 and replace all digits right to it by 0. So, our required number will be 220.

Tricky Cases: 

  • If the first odd digit in N is 9, then we must replace the digit immediately to the left of that odd digit with the next even digit. For example, if N=44934, then smallest number=46000.
  • Another tricky case is when the first odd digit is 9 and the digit directly to the left of the first odd digit is 8. In this case, we must replace the digit directly to the left of first odd digit by 0 and the digit left to this digit by next even digit, and keep doing this until we find a digit other than 8. For example, if N=86891, then Y=88000. Finally, if all digits to the left continue to be 8 until we reach the leftmost digit, or if the first digit of N is 9, then we must add the smallest non-zero even digit (i.e. 2) as a new digit on the left. For example, if N=891 or N=910, then Y=2000.


Below is the implementation of the efficient approach: 
 

C++ 
// CPP program to print the smallest
// integer not less than N with all even digits
#include <bits/stdc++.h>
using namespace std;

// function to return the answer when the
// first odd digit is 9
int trickyCase(string s, int index)
{

    int index1 = -1;

    // traverse towards the left to find the non-8 digit
    for (int i = index - 1; i >= 0; i--) {
        // index digit
        int digit = s[i] - '0';

        // if digit is not 8, then break
        if (digit != 8) {
            index1 = i;
            break;
        }
    }
    // if on the left side of the '9', no 8
    // is found then we return by adding a 2 and 0's
    if (index1 == -1)
        return 2 * pow(10, s.length());

    int num = 0;

    // till non-8 digit add all numbers
    for (int i = 0; i < index1; i++)
        num = num * 10 + (s[i] - '0');

    // if non-8 is even or odd than add the next even.
    if (s[index1] % 2 == 0)
        num = num * 10 + (s[index1] - '0' + 2);
    else
        num = num * 10 + (s[index1] - '0' + 1);

    // add 0 to right of 9
    for (int i = index1 + 1; i < s.length(); i++)
        num = num * 10;

    return num;
}

// function to return the smallest number
// with all digits even
int smallestNumber(int n)
{
    int num = 0;
    string s = "";

    int duplicate = n;
    // convert the number to string to
    // perform operations
    while (n) {
        s = char(n % 10 + 48) + s;
        n /= 10;
    }

    int index = -1;

    // find out the first odd number
    for (int i = 0; i < s.length(); i++) {
        int digit = s[i] - '0';
        if (digit & 1) {
            index = i;
            break;
        }
    }

    // if no odd numbers are there, than n is the answer
    if (index == -1)
        return duplicate;

    // if the odd number is 9,
    // than tricky case handles it
    if (s[index] == '9') {
        num = trickyCase(s, index);
        return num;
    }

    // add all digits till first odd
    for (int i = 0; i < index; i++)
        num = num * 10 + (s[i] - '0');

    // increase the odd digit by 1
    num = num * 10 + (s[index] - '0' + 1);

    // add 0 to the right of the odd number
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10;

    return num;
}

// Driver Code
int main()
{
    int N = 2397;
    cout << smallestNumber(N);

    return 0;
}
Java 
// Java program to print the 
// smallest integer not less
// than N with all even digits
import java.io.*;
import java.util.*;
import java.lang.*;

class GFG
{
// function to return 
// the answer when the
// first odd digit is 9
static int trickyCase(String s, 
                      int index)
{

    int index1 = -1;

    // traverse towards the left 
    // to find the non-8 digit
    for (int i = index - 1; i >= 0; i--) 
    {
        // index digit
        int digit = s.charAt(i) - '0';

        // if digit is not 8,
        // then break
        if (digit != 8) 
        {
            index1 = i;
            break;
        }
    }
    
    // if on the left side of the 
    // '9', no 8 is found then we
    // return by adding a 2 and 0's
    if (index1 == -1)
        return 2 * (int)Math.pow(10, s.length());

    int num = 0;

    // till non-8 digit 
    // add all numbers
    for (int i = 0; i < index1; i++)
        num = num * 10 + (s.charAt(i) - '0');

    // if non-8 is even or odd 
    // than add the next even.
    if (s.charAt(index1) % 2 == 0)
        num = num * 10 + 
            (s.charAt(index1) - '0' + 2);
    else
        num = num * 10 + 
            (s.charAt(index1) - '0' + 1);

    // add 0 to right of 9
    for (int i = index1 + 1; 
            i < s.length(); i++)
        num = num * 10;

    return num;
}

// function to return 
// the smallest number
// with all digits even
static int smallestNumber(int n)
{
    int num = 0;
    String s = "";

    int duplicate = n;
    
    // convert the number to 
    // string to perform operations
    while (n > 0) 
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }

    int index = -1;

    // find out the
    // first odd number
    for (int i = 0; i < s.length(); i++) 
    {
        int digit = s.charAt(i) - '0';
        int val = digit & 1;
        if (val == 1)
        {
            index = i;
            break;
        }
    }

    // if no odd numbers are there,
    // than n is the answer
    if (index == -1)
        return duplicate;

    // if the odd number is 9,
    // than tricky case handles it
    if (s.charAt(index) == '9') 
    {
        num = trickyCase(s, index);
        return num;
    }

    // add all digits till first odd
    for (int i = 0; i < index; i++)
        num = num * 10 + 
             (s.charAt(i) - '0');

    // increase the 
    // odd digit by 1
    num = num * 10 + 
        (s.charAt(index) - '0' + 1);

    // add 0 to the right
    // of the odd number
    for (int i = index + 1; 
             i < s.length(); i++)
        num = num * 10;

    return num;
}

// Driver Code
public static void main(String args[])
{
    int N = 2397;
    System.out.print(smallestNumber(N));
}
}

// This code is contributed
// by Akanksha rai(Abby_akku)
Python3 
# Python3 program to print the smallest
# integer not less than N with all even digits

# Function to return the answer when the
# first odd digit is 9
def trickyCase(s, index):
    index1 = -1;

    # traverse towards the left to find 
    # the non-8 digit
    for i in range(index - 1, -1, -1): 

        # index digit
        digit = s[i] - '0';

        # if digit is not 8, then break
        if (digit != 8):
            index1 = i;
            break;
    
    # if on the left side of the '9', 
    # no 8 is found then we return by 
    # adding a 2 and 0's
    if (index1 == -1):
        return 2 * pow(10, len(s));

    num = 0;

    # till non-8 digit add all numbers
    for i in range(index1):
        num = num * 10 + (s[i] - '0');

    # if non-8 is even or odd 
    # than add the next even.
    if (s[index1] % 2 == 0):
        num = num * 10 + (s[index1] - '0' + 2);
    else:
        num = num * 10 + (s[index1] - '0' + 1);

    # add 0 to right of 9
    for i in range(index1 + 1, len(s)):
        num = num * 10;

    return num;

# function to return the smallest 
# number with all digits even
def smallestNumber(n):

    num = 0;
    s = "";

    duplicate = n;
    
    # convert the number to string to 
    # perform operations
    while (n): 
        s = chr(n % 10 + 48) + s;
        n = int(n / 10);

    index = -1;

    # find out the first odd number
    for i in range(len(s)):
        digit = ord(s[i]) - ord('0');
        if (digit & 1):
            index = i;
            break;

    # if no odd numbers are
    # there, than n is the answer
    if (index == -1):
        return duplicate;

    # if the odd number is 9, than 
    # tricky case handles it
    if (s[index] == '9'):
        num = trickyCase(s, index);
        return num;

    # add all digits till first odd
    for i in range(index):
        num = num * 10 + ord(s[i]) - ord('0');

    # increase the odd digit by 1
    num = num * 10 + (ord(s[index]) - 
                      ord('0') + 1);

    # add 0 to the right of the odd number
    for i in range(index + 1, len(s)):
        num = num * 10;

    return num;

# Driver Code
N = 2397;
print(smallestNumber(N));

# This code is contributed
# by mits
C# 
// C# program to print the smallest integer 
// not less than N with all even digits
using System;

class GFG
{
// function to return the answer when 
// the first odd digit is 9
static int trickyCase(string s, 
                      int index)
{
    int index1 = -1;

    // traverse towards the left 
    // to find the non-8 digit
    for (int i = index - 1; i >= 0; i--) 
    {
        // index digit
        int digit = s[i] - '0';

        // if digit is not 8, then break
        if (digit != 8) 
        {
            index1 = i;
            break;
        }
    }
    
    // if on the left side of the 
    // '9', no 8 is found then we
    // return by adding a 2 and 0's
    if (index1 == -1)
        return 2 * (int)Math.Pow(10, s.Length);

    int num = 0;

    // till non-8 digit add all numbers
    for (int i = 0; i < index1; i++)
        num = num * 10 + (s[i] - '0');

    // if non-8 is even or odd 
    // than add the next even.
    if (s[index1] % 2 == 0)
        num = num * 10 + 
            (s[index1] - '0' + 2);
    else
        num = num * 10 + 
             (s[index1] - '0' + 1);

    // add 0 to right of 9
    for (int i = index1 + 1; 
            i < s.Length; i++)
        num = num * 10;

    return num;
}

// function to return the smallest number
// with all digits even
static int smallestNumber(int n)
{
    int num = 0;
    string s = "";

    int duplicate = n;
    
    // convert the number to 
    // string to perform operations
    while (n > 0) 
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }

    int index = -1;

    // find out the first odd number
    for (int i = 0; i < s.Length; i++) 
    {
        int digit = s[i] - '0';
        int val = digit & 1;
        if (val == 1)
        {
            index = i;
            break;
        }
    }

    // if no odd numbers are there,
    // than n is the answer
    if (index == -1)
        return duplicate;

    // if the odd number is 9,
    // than tricky case handles it
    if (s[index] == '9') 
    {
        num = trickyCase(s, index);
        return num;
    }

    // add all digits till first odd
    for (int i = 0; i < index; i++)
        num = num * 10 + 
            (s[i] - '0');

    // increase the odd digit by 1
    num = num * 10 + 
        (s[index] - '0' + 1);

    // add 0 to the right of the odd number
    for (int i = index + 1; 
            i < s.Length; i++)
        num = num * 10;

    return num;
}

// Driver Code
public static void Main()
{
    int N = 2397;
    Console.Write(smallestNumber(N));
}
}

// This code is contributed
// by Akanksha Rai
?>
PHP 
<?php
// PHP program to print 
// the smallest integer
// not less than N with
// all even digits

// function to return
// the answer when the
// first odd digit is 9
function trickyCase($s, $index)
{
    $index1 = -1;

    // traverse towards the 
    // left to find the 
    // non-8 digit
    for ($i = $index - 1; 
         $i >= 0; $i--) 
    {
        // index digit
        $digit = $s[$i] - '0';

        // if digit is not
        // 8, then break
        if ($digit != 8)
        {
            $index1 = $i;
            break;
        }
    }
    
    // if on the left side 
    // of the '9', no 8
    // is found then we 
    // return by adding a 2 
    // and 0's
    if ($index1 == -1)
        return 2 * pow(10, 
              strlen($s));

    $num = 0;

    // till non-8 digit
    // add all numbers
    for ($i = 0; $i < $index1; $i++)
        $num = $num * 10 + 
              ($s[$i] - '0');

    // if non-8 is even or 
    // odd than add the next even.
    if ($s[$index1] % 2 == 0)
        $num = $num * 10 + 
              ($s[$index1] - '0' + 2);
    else
        $num = $num * 10 + 
              ($s[$index1] - '0' + 1);

    // add 0 to right of 9
    for ($i = $index1 + 1; 
         $i < strlen($s); $i++)
        $num = $num * 10;

    return $num;
}

// function to return
// the smallest number
// with all digits even
function smallestNumber($n)
{
    $num = 0;
    $s = "";

    $duplicate = $n;
    
    // convert the number 
    // to string to perform
    // operations
    while ($n) 
    {
        $s = chr($n % 10 + 48) . $s;
        $n = (int)($n / 10);
    }

    $index = -1;

    // find out the 
    // first odd number
    for ($i = 0; 
         $i < strlen($s); $i++)
    {
        $digit = $s[$i] - '0';
        if ($digit & 1)
        {
            $index = $i;
            break;
        }
    }

    // if no odd numbers are
    // there, than n is the answer
    if ($index == -1)
        return $duplicate;

    // if the odd number 
    // is 9, than tricky 
    // case handles it
    if ($s[$index] == '9') 
    {
        $num = trickyCase($s, $index);
        return $num;
    }

    // add all digits
    // till first odd
    for ($i = 0; $i < $index; $i++)
        $num = $num * 10 + 
              ($s[$i] - '0');

    // increase the 
    // odd digit by 1
    $num = $num * 10 + 
          ($s[$index] - '0' + 1);

    // add 0 to the right 
    // of the odd number
    for ($i = $index + 1; 
         $i < strlen($s); $i++)
        $num = $num * 10;

    return $num;
}

// Driver Code
$N = 2397;
echo smallestNumber($N);

// This code is contributed
// by mits
?>
JavaScript 
<script>
    // Javascript program to print the smallest integer
    // not less than N with all even digits
    
    // function to return the answer when
    // the first odd digit is 9
    function trickyCase(s, index)
    {
        let index1 = -1;

        // traverse towards the left
        // to find the non-8 digit
        for (let i = index - 1; i >= 0; i--)
        {
            // index digit
            let digit = s[i].charCodeAt() - '0'.charCodeAt();

            // if digit is not 8, then break
            if (digit != 8)
            {
                index1 = i;
                break;
            }
        }

        // if on the left side of the
        // '9', no 8 is found then we
        // return by adding a 2 and 0's
        if (index1 == -1)
            return 2 * Math.pow(10, s.length);

        let num = 0;

        // till non-8 digit add all numbers
        for (let i = 0; i < index1; i++)
            num = num * 10 + (s[i].charCodeAt() - '0'.charCodeAt());

        // if non-8 is even or odd
        // than add the next even.
        if (s[index1].charCodeAt() % 2 == 0)
            num = num * 10 +
                (s[index1].charCodeAt() - '0'.charCodeAt() + 2);
        else
            num = num * 10 +
                 (s[index1].charCodeAt() - '0'.charCodeAt() + 1);

        // add 0 to right of 9
        for (let i = index1 + 1;
                i < s.length; i++)
            num = num * 10;

        return num;
    }

    // function to return the smallest number
    // with all digits even
    function smallestNumber(n)
    {
        let num = 0;
        let s = "";

        let duplicate = n;

        // convert the number to
        // string to perform operations
        while (n > 0)
        {
            s = String.fromCharCode(n % 10 + 48) + s;
            n = parseInt(n / 10, 10);
        }

        let index = -1;

        // find out the first odd number
        for (let i = 0; i < s.length; i++)
        {
            let digit = s[i].charCodeAt() - '0'.charCodeAt();
            let val = digit & 1;
            if (val == 1)
            {
                index = i;
                break;
            }
        }

        // if no odd numbers are there,
        // than n is the answer
        if (index == -1)
            return duplicate;

        // if the odd number is 9,
        // than tricky case handles it
        if (s[index] == '9')
        {
            num = trickyCase(s, index);
            return num;
        }

        // add all digits till first odd
        for (let i = 0; i < index; i++)
            num = num * 10 +
                (s[i].charCodeAt() - '0'.charCodeAt());

        // increase the odd digit by 1
        num = num * 10 +
            (s[index].charCodeAt() - '0'.charCodeAt() + 1);

        // add 0 to the right of the odd number
        for (let i = index + 1;
                i < s.length; i++)
            num = num * 10;

        return num;
    }
    
    let N = 2397;
    document.write(smallestNumber(N));
    
    // This code is contributed by divyeshrabadiya07.
</script>

Output
2400

Time Complexity: O(log10N)
Auxiliary Space: O(log10N)


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