Maximum possible number with the given operation

Last Updated : 19 Mar, 2022

Given a positive integer N, the task is to convert this integer to the maximum possible integer without leading zeroes by changing the digits. A digit X can only be changed into a digit Y if X + Y = 9.
Examples:

Input: N = 42
Output: 57
Change 4 -> 5 and 2 -> 7.
Input: N = 1
Output: 8

Approach: Only the digits which are greater than or equal to 5 need to be changed as changing the digits which are less than 5 will result in a larger number. After all the required digits have been updated, check whether the resultant number has a leading zero, if yes then change it to a 9.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum possible
// integer that can be obtained from the
// given integer after performing
// the given operations
string maxInt(string str)
{
    // For every digit
    for (int i = 0; i < str.length(); i++) {
 
        // Digits greater than or equal to 5
        // need not to be changed as changing
        // them will lead to a smaller number
        if (str[i] < '5') {
            str[i] = ('9' - str[i]) + '0';
        }
    }
 
    // The resulting integer
    // cannot have leading zero
    if (str[0] == '0')
        str[0] = '9';
 
    return str;
}
 
// Driver code
int main()
{
    string str = "42";
 
    cout << maxInt(str);
 
    return 0;
}

Java

// Java implementation of the approach 
class GFG 
{
 
    // Function to return the maximum possible 
    // integer that can be obtained from the 
    // given integer after performing 
    // the given operations 
    static String maxInt(char str[]) 
    { 
        // For every digit 
        for (int i = 0; i < str.length; i++)
        { 
     
            // Digits greater than or equal to 5 
            // need not to be changed as changing 
            // them will lead to a smaller number 
            if (str[i] < '5')
            { 
                str[i] = (char)(('9' - str[i]) + '0'); 
            } 
        } 
     
        // The resulting integer 
        // cannot have leading zero 
        if (str[0] == '0') 
            str[0] = '9'; 
     
        String str2 = new String(str);
        return str2; 
    } 
     
    // Driver code 
    public static void main (String[] args) 
    { 
        String str = "42"; 
     
        System.out.println(maxInt(str.toCharArray())); 
    } 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach 
 
# Function to return the maximum possible 
# integer that can be obtained from the 
# given integer after performing 
# the given operations 
def maxInt(string): 
    string2 = ""
     
    # For every digit 
    for i in range(0, len(string)): 
 
        # Digits greater than or equal to 5 
        # need not to be changed as changing 
        # them will lead to a smaller number 
        if (string[i] < '5'):
            string2 += str((ord('9') - ord(string[i])))
        else:
            string2 += str(string[i])
             
    # The resulting integer 
    # cannot have leading zero 
    if (string2[0] == '0'): 
        string2[0] = '9'
 
    return string2 
 
# Driver code 
if __name__ == '__main__': 
 
    string = "42"
    print(maxInt(string)) 
 
# This code is contributed by ashutosh450

C#

// C# implementation of the approach
using System; 
 
class GFG 
{
 
    // Function to return the maximum possible 
    // integer that can be obtained from the 
    // given integer after performing 
    // the given operations 
    static String maxInt(char []str) 
    { 
        // For every digit 
        for (int i = 0; i < str.Length; i++)
        { 
     
            // Digits greater than or equal to 5 
            // need not to be changed as changing 
            // them will lead to a smaller number 
            if (str[i] < '5')
            { 
                str[i] = (char)(('9' - str[i]) + '0'); 
            } 
        } 
     
        // The resulting integer 
        // cannot have leading zero 
        if (str[0] == '0') 
            str[0] = '9'; 
     
        String str2 = new String(str);
        return str2; 
    } 
     
    // Driver code 
    public static void Main (String []args) 
    { 
        String str = "42"; 
     
        Console.WriteLine(maxInt(str.ToCharArray())); 
    } 
}
 
// This code is contributed by Arnab Kundu

Javascript

<script>
// Javascript implementation of the
// above approach
 
// Function to return the maximum possible
// integer that can be obtained from the
// given integer after performing
// the given operations
function maxInt(str)
{
    // For every digit
    var str2 = "";
    for (var i = 0; i < str.length; i++) {
   
        // Digits greater than or equal to 5
        // need not to be changed as changing
        // them will lead to a smaller number
        if (str[i] < '5') {
            var l = ('9'.charCodeAt(0) - str[i].charCodeAt(0)) + '0'.charCodeAt(0);
            str2 = str2.concat(String.fromCharCode(l));
        }
    }
   
    // The resulting integer
    // cannot have leading zero
    if (str2[0] == '0')
        str2[0] = '9';
   
    return str2;
}
 
// Driver code
var str = "42";
document.write(maxInt(str)) 
 
// This code is contributed by ShubhamSingh10
</script>

Output:

Time Complexity: O(|str|)

Auxiliary Space: O(1)

Maximize the number of local maxima's with minimum operations

spp____

Improve

Article Tags :

Practice Tags :

Mathematical

Maximum possible number with the given operation

C++

Java

Python3

C#

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?