Minimum possible number with the given operation

Given a positive integer N, the task is to convert this integer to the minimum 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 = 589
Output: 410
Change 5 -> 4, 8 -> 1 and 9 -> 0



Input: N = 934
Output: 934
934 cannot be minimised.

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum possible
// integer that can be obtained from the
// given integer after performing
// the given operations
string minInt(string str)
{
    // For every digit
    for (int i = 0; i < str.length(); i++) {
  
        // Digits less than 5 need not to be
        // changed as changing them will
        // lead to a larger 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 = "589";
  
    cout << minInt(str);
  
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
   
# Function to return the minimum possible
# integer that can be obtained from the
# given integer after performing
# the given operations
def minInt(str1):
      
    # For every digit
    for i in range(len(str1)):
  
        # Digits less than 5 need not to be
        # changed as changing them will
        # lead to a larger number
        if (str1[i] >= 5):
            str1[i] = (9 - str1[i])
  
    # The resulting integer
    # cannot have leading zero
    if (str1[0] == 0):
        str1[0] = 9
          
    temp = ""
  
    for i in str1:
        temp += str(i)
  
    return temp
  
# Driver code
str1 = "589"
str1 = [int(i) for i in str1]
  
print(minInt(str1))
  
# This code is contributed by Mohit Kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach 
using System;
  
class GFG
{
      
    // Function to return the minimum possible 
    // integer that can be obtained from the 
    // given integer after performing 
    // the given operations 
    static string minInt(char []str) 
    
        // For every digit 
        for (int i = 0; i < str.Length; i++)
        
      
            // Digits less than 5 need not to be 
            // changed as changing them will 
            // lead to a larger number 
            if ((int)str[i] >= (int)('5')) 
            
                str[i] = (char)(((int)('9') - 
                                 (int)(str[i])) + 
                                 (int)('0')); 
            
        
      
        // The resulting integer 
        // cannot have leading zero 
        if (str[0] == '0'
            str[0] = '9'
      
        string s = new string(str);
        return s; 
    
      
    // Driver code 
    static public void Main ()
    
        string str = "589"
        Console.WriteLine(minInt(str.ToCharArray())); 
    
}
  
// This code is contributed by AnkitRai01

chevron_right


Output:

410


My Personal Notes arrow_drop_up

Budding Web DeveloperKeen learnerAverage CoderDancer&Social Activist

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29, AnkitRai01



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.