Minimum number closest to N made up of odd digits only

Given an integer N, the task is to find the number closest to N having only odd digits. If more than one such number exists, print the minimum.

Examples:

Input: N = 725
Output: 719
Explanation:
Numbers 719 and 731 consists of odd digits only and are closest to 725. Since 719 is the minimum, it is the required answer.

Input : N = 111
Output: 111

Naive Approach: The simplest approach to solve the problem is to iterate and find the closest numbers smaller and greater than N, having only odd numbers as its digits. Follow the steps below to solve the problem:

1. If all the digits of N are odd, then print N as the answer.
2. Find the closest smaller and greater numbers and return the one having the least difference with N. If both are found to be equidistant from N, print the smaller number.

Time Complexity: O(10(len(N) – 1))
Auxiliary Space:O(1)

Efficient Approach: Follow the steps below to optimize the above approach:

1. If all the digits of N are odd, then return N.
2. Calculate the closest smaller number to N efficiently by the following steps:
• Find the position of first even digit in N from left to right, i.e, pos.
• If the first even digit is 0, then replace 0 with 9 and iterate over all preceding digits and for every digit, if found to be 1, replace it with 9. Otherwise, decrease the digit by 2 and return the number.
• If no preceding digits are found to be exceeding 1, then replace the most significant digit with 0.
• If the first even digit is non-zero, then decrease that digit by 1. Replace all the succeeding digits with 9.
3. Calculate the greater number closest to N by the steps below:
• Find the position of first even digit i.e pos.
• Increment the digits at pos by 1.
• Iterate over the succeeding digits and increment them by 1.
4. Compare the absolute difference of the respective closest numbers obtained with N and print the one having least difference.
5. If both the differences are found to be equal, print the smaller number.

Below is the implementation of the above approach:

C++

 #include #include #include #include   using namespace std;   // Function to return the smaller // number closest to N made up of // only odd digits string closest_smaller(int N) {     string strN = to_string(N);     vector l(strN.length());     for (int i = 0; i < strN.length(); i++) {         l[i] = strN[i] - '0';     }     int length = l.size();     int pos = -1;     for (int i = 0; i < length; i++) {         if (l[i] % 2 == 0) {             pos = i;             break;         }     }     if (l[pos] == 0) {         l[pos] = 9;         for (int i = pos - 1; i >= 0; i--) {             if (l[i] == 1) {                 if (i == 0) {                     l[i] = 0;                     break;                 }                 else {                     l[i] = 9;                 }             }             else {                 l[i] -= 2;                 break;             }         }     }     else {         l[pos] -= 1;     }     for (int i = pos + 1; i < length; i++) {         l[i] = 9;     }     if (l[0] == 0) {         l.erase(l.begin());     }     string result;     for (int i = 0; i < l.size(); i++) {         result += to_string(l[i]);     }     return result; }   // Function to return the greater // number closest to N made up of // only odd digits string closest_greater(int N) {     string strN = to_string(N);     vector l(strN.length());     for (int i = 0; i < strN.length(); i++) {         l[i] = strN[i] - '0';     }     int length = l.size();     int pos = -1;     for (int i = 0; i < length; i++) {         if (l[i] % 2 == 0) {             pos = i;             break;         }     }     l[pos] += 1;     for (int i = pos + 1; i < length; i++) {         l[i] = 1;     }     string result;     for (int i = 0; i < l.size(); i++) {         result += to_string(l[i]);     }     return result; }   // Function to check if all // digits of N are odd or not bool check_all_digits_odd(int N) {     string strN = to_string(N);     vector l(strN.length());     for (int i = 0; i < strN.length(); i++) {         l[i] = strN[i] - '0';     }     int length = l.size();     int pos = -1;     for (int i = 0; i < length; i++) {         if (l[i] % 2 == 0) {             pos = i;             break;         }     }     if (pos == -1) {         return true;     }     return false; }   // Function to return the // closest number to N // having odd digits only void closestNumber(int N) {     if (check_all_digits_odd(N)) {           cout << N << endl;     }     else {         int l = stoi(closest_smaller(N));         int r = stoi(closest_greater(N));         if (abs(N - l) <= abs(N - r)) {             cout << l << endl;         }         else {             cout << r << endl;         }     } }   // Driver code int main() {     int N = 110;     closestNumber(N);     return 0; }

Java

 // Java program for the above approach   import java.util.*;   public class GFG {           // Function to return the smaller     // number closest to N made up of     // only odd digits     static String ClosestSmaller(String n) {         char[] digits = n.toCharArray();         int length = digits.length;               int pos = -1;               for (int i = 0; i < length; i++) {             if (digits[i] % 2 == 0) {                 pos = i;                 break;             }         }               if (digits[pos] == '0') {             digits[pos] = '9';                   for (int i = pos - 1; i >= 0; i--) {                 if (digits[i] == '1') {                     if (i == 0) {                         digits[i] = '0';                         break;                     }                           digits[i] = '9';                 } else {                     digits[i] = (char) (digits[i] - 2);                     break;                 }             }         } else {             digits[pos] = (char) (digits[pos] - 1);         }               for (int i = pos + 1; i < length; i++) {             digits[i] = '9';         }               if (digits[0] == '0') {             digits = Arrays.copyOfRange(digits, 1, digits.length);         }               String result = new String(digits);         return result;     }           // Function to return the greater     // number closest to N made up of     // only odd digits     static String ClosestGreater(String n) {         char[] digits = n.toCharArray();         int length = digits.length;               int pos = -1;               for (int i = 0; i < length; i++) {             if (digits[i] % 2 == 0) {                 pos = i;                 break;             }         }               digits[pos] = (char) (digits[pos] + 1);               for (int i = pos + 1; i < length; i++) {             digits[i] = '1';         }               String result = new String(digits);         return result;     }           // Function to check if all     // digits of N are odd or not     static boolean CheckAllDigitsOdd(String n) {         char[] digits = n.toCharArray();         int length = digits.length;               int pos = -1;               for (int i = 0; i < length; i++) {             if ((digits[i] - '0') % 2 == 0) {                 pos = i;                 break;             }         }               if (pos == -1)             return true;               return false;     }               // Function to return the     // closest number to N     // having odd digits only     static void ClosestNumber(String n) {         if (CheckAllDigitsOdd(n)) {             System.out.println(n);         } else {             int l = Integer.parseInt(ClosestSmaller(n));                   int r = Integer.parseInt(ClosestGreater(n));                   if (Math.abs(Integer.parseInt(n) - l)                     < Math.abs(Integer.parseInt(n) - r)) {                 System.out.println(l);             } else {                 System.out.println(r);             }         }     }     // Driver code     public static void main(String[] args) {         String n = "110";         ClosestNumber(n);     } }   // This code is contributed by princekumaras

Python

 # Python program to implement # the above approach   # Function to return the smaller # number closest to N made up of # only odd digits def closest_smaller(N):       N = str(N)       l = list(map(int, N))     length = len(l)           # Stores the position of     # first even digit of N     pos = -1       # Iterate through each digit of N     for i in range(length):                   # Check for even digit.         if l[i] % 2 == 0:             pos = i             break       # If the first even digit is 0     if l[pos] == 0:           # Replace 0 with 9         l[pos] = 9           # Iterate over preceding         # digits         for i in range(pos - 1, -1, -1):               # If current digit is 1             if l[i] == 1:                   # Check if it is the                 # first digit or not                 if i == 0:                                           # Append leading 0's                     l[i] = 0                     break                                       # Otherwise, replace by 9                 l[i] = 9                               # Otherwise             else:                   # Decrease its value by 2                 l[i] -= 2                 break       # If the first even digit exceeds 0     else:                   # Reduce the digit by 1         l[pos] -= 1       # Replace all succeeding digits by 9     for i in range(pos + 1, length):         l[i] = 9       # Remove leading 0s     if l[0] == 0:         l.pop(0)                 result = ''.join(map(str, l))     return result       # Function to return the greater # number closest to N made up of # only odd digits def closest_greater(N):       N = str(N)       l = list(map(int, N))     length = len(l)       # Stores the position of     # first even digit of N     pos = -1       # Iterate over each digit     # of N     for i in range(length):                   # If even digit is found         if l[i] % 2 == 0:             pos = i             break                   # Increase value of first     # even digit by 1     l[pos] += 1       for i in range(pos + 1, length):         l[i] = 1       result = ''.join(map(str, l))     return result   # Function to check if all # digits of N are odd or not def check_all_digits_odd(N):       N = str(N)       l = list(map(int, N))     length = len(l)       # Stores the position of     # first even digit of N     pos = -1       # Iterating over each digit     # of N     for i in range(length):                   # If even digit is found         if l[i] % 2 == 0:             pos = i             break                   # If no even digit is found     if pos == -1:         return True     return False   # Function to return the # closest number to N # having odd digits only def closestNumber(N):           # If all digits of N are odd     if check_all_digits_odd(N):         print(N)     else:                   # Find smaller number         # closest to N         l = int(closest_smaller(N))                   # Find greater number         # closest to N         r = int(closest_greater(N))                   # Print the number with least         # absolute difference         if abs(N - l) <= abs(N - r):             print(l)         else:             print(r)   # Driver Code. if __name__ == '__main__':           N = 110     closestNumber(N)

C#

 using System;   class GFG {     // Function to return the smaller   // number closest to N made up of   // only odd digits   static string ClosestSmaller(string n)   {     char[] digits = n.ToCharArray();     int length = digits.Length;       // Stores the position of     // first even digit of N     int pos = -1;       // Iterate through each digit of N     for (int i = 0; i < length; i++) {       // Check for even digit.       if (digits[i] % 2 == 0) {         pos = i;         break;       }     }       // If the first even digit is 0     if (digits[pos] == '0') {       // Replace 0 with 9       digits[pos] = '9';         // Iterate over preceding       // digits       for (int i = pos - 1; i >= 0; i--) {         // If current digit is 1         if (digits[i] == '1') {           // Check if it is the           // first digit or not           if (i == 0) {             // Append leading 0's             digits[i] = '0';             break;           }             // Otherwise, replace by 9           digits[i] = '9';         }         else {           // Decrease its value by 2           digits[i] = (char)(digits[i] - 2);           break;         }       }     }     else {       // If the first even digit exceeds 0,       // reduce the digit by 1       digits[pos] = (char)(digits[pos] - 1);     }       // Replace all succeeding digits by 9     for (int i = pos + 1; i < length; i++) {       digits[i] = '9';     }       // Remove leading 0s     if (digits[0] == '0') {       digits = digits[1..];     }       string result = new string(digits);     return result;   }     // Function to return the greater   // number closest to N made up of   // only odd digits   static string ClosestGreater(string n)   {     char[] digits = n.ToCharArray();     int length = digits.Length;       // Stores the position of     // first even digit of N     int pos = -1;       // Iterate over each digit     // of N     for (int i = 0; i < length; i++) {       // If even digit is found       if (digits[i] % 2 == 0) {         pos = i;         break;       }     }       // Increase value of first     // even digit by 1     digits[pos] = (char)(digits[pos] + 1);       for (int i = pos + 1; i < length; i++) {       digits[i] = '1';     }       string result = new string(digits);     return result;   }     // Function to check if all   // digits of N are odd or not   static bool CheckAllDigitsOdd(string n)   {     char[] digits = n.ToCharArray();     int length = digits.Length;       // Stores the position of     // first even digit of N     int pos = -1;       // Iterating over each digit     // of N     for (int i = 0; i < length; i++) {       // If even digit is found       if ((digits[i] - '0') % 2 == 0) {         pos = i;         break;       }     }       // If no even digit is found     if (pos == -1)       return true;       return false;   }     // Function to return the   // closest number to N   // having odd digits only   static void ClosestNumber(string n)   {     // If all digits of N are odd     if (CheckAllDigitsOdd(n)) {       Console.WriteLine(n);     }     else {       // Find smaller number       // closest to N       int l = Convert.ToInt32(ClosestSmaller(n));         // Find greater number       // closest to N       int r = Convert.ToInt32(ClosestGreater(n));         // Print the number with least       // absolute difference       if (Math.Abs(Convert.ToInt32(n) - l)           < Math.Abs(Convert.ToInt32(n) - r)) {         Console.WriteLine(l);       }       else {         Console.WriteLine(r);       }     }   }     static void Main(string[] args)   {     // Driver Code.     string n = "110";     ClosestNumber(n);   } }

Javascript

 // JavaScript program to implement // the above approach   // Function to return the smaller // number closest to N made up of // only odd digits function closest_smaller(N) {       N = N.toString();       let l = N.split('').map(Number);     let length = l.length;       // Stores the position of     // first even digit of N     let pos = -1;       // Iterate through each digit of N     for (let i = 0; i < length; i++) {           // Check for even digit.         if (l[i] % 2 === 0) {             pos = i;             break;         }     }       // If the first even digit is 0     if (l[pos] === 0) {           // Replace 0 with 9         l[pos] = 9;           // Iterate over preceding         // digits         for (let i = pos - 1; i >= 0; i--) {               // If current digit is 1             if (l[i] === 1) {                   // Check if it is the                 // first digit or not                 if (i === 0) {                       // Append leading 0's                     l[i] = 0;                     break;                 }                   // Otherwise, replace by 9                 l[i] = 9;               } else {                   // Decrease its value by 2                 l[i] -= 2;                 break;             }         }     } else {           // If the first even digit exceeds 0,         // reduce the digit by 1         l[pos] -= 1;     }       // Replace all succeeding digits by 9     for (let i = pos + 1; i < length; i++) {         l[i] = 9;     }       // Remove leading 0s     if (l[0] === 0) {         l.shift();     }       let result = l.join('');     return result; }   // Function to return the greater // number closest to N made up of // only odd digits function closest_greater(N) {       N = N.toString();       let l = N.split('').map(Number);     let length = l.length;       // Stores the position of     // first even digit of N     let pos = -1;       // Iterate over each digit     // of N     for (let i = 0; i < length; i++) {           // If even digit is found         if (l[i] % 2 === 0) {             pos = i;             break;         }     }       // Increase value of first     // even digit by 1     l[pos] += 1;       for (let i = pos + 1; i < length; i++) {         l[i] = 1;     }       let result = l.join('');     return result; }   // Function to check if all // digits of N are odd or not function check_all_digits_odd(N) {       N = N.toString();       let l = N.split('').map(Number);     let length = l.length;       // Stores the position of     // first even digit of N     let pos = -1;       // Iterating over each digit     // of N     for (let i = 0; i < length; i++) {           // If even digit is found         if (l[i] % 2 == 0) {             pos = i;             break;         }     }       // If no even digit is found     if (pos == -1)         return true     return false       }   // Function to return the // closest number to N // having odd digits only function closestNumber(N) {           // If all digits of N are odd     if (check_all_digits_odd(N))         console.log(N)     else {         // Find smaller number         // closest to N         let l = parseInt(closest_smaller(N))           // Find greater number         // closest to N         let r = parseInt(closest_greater(N))           // Print the number with least         // absolute difference         if (Math.abs(N - l) < Math.abs(N - r))             console.log(l)         else             console.log(r)     } }   // Driver Code. let N = 110 closestNumber(N)     // by phasing17

Output:

111

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