Maximum 69 Number

Published: Nov 7, 2022

Introduction

The given number consists of only 6 and/or 9, and only one character is allowed to change. Given that, find 6 from left, replace the first 6 to 9. Only point we should care about is whether 6 exists or not. If not, the given string is all 9s and no need to change.

Problem Description

You are given a positive integer num consisting only of digits 6 and 9.

Return the maximum number you can get by changing at most one digit (6 becomes 9, and 9 becomes 6).

Constraints:

• 1 <= num <= 10**4
• num consists of only 6 and 9 digits.

https://leetcode.com/problems/maximum-69-number/

Examples

Example 1
Input: num = 9669
Output: 9969

Example 2
Input: num = 9996
Output: 9999

Example 3
Input: num = 9999
Output: 9999

Analysis

It’s easy to check every digit if the number is string. The solution here starts converting the given number to string. Once the index of the first 6 is found, substring concatenation creates the answer. Lastly, convert the result string to integer to meet with the requirement.

Solution

class Maximum69Number {
public:
    int maximum69Number (int num) {
        std::string s = std::to_string(num);
        std::size_t found = s.find("6");
        if (found != std::string::npos) {
            s.replace(found, 1, "9");
        }
        return std::stoi(s);
    }
};

class Maximum69Number {
    public int maximum69Number (int num) {
        String s = String.valueOf(num);
        int found = s.indexOf("6");
        if (found >= 0) {
            s = s.replaceFirst("6", "9");
        }
        return Integer.parseInt(s);
    }
}

/**
 * @param {number} num
 * @return {number}
 */
var maximum69Number  = function(num) {
        let s = num.toString();
        let found = s.indexOf("6");
        if (found >= 0) {
            s = s.replace('6', '9');
        }
        return parseInt(s);
    };

class Maximum69Number:
    def maximum69Number (self, num: int) -> int:
        s = str(num)
        if '6' in s:
            idx = s.index('6')
            s = s[:idx] + '9' + s[idx+1:]
        return int(s)

# @param {Integer} num
# @return {Integer}
def maximum69_number (num)
  s = num.to_s
  if s.count('6') > 0
    idx = s.index('6')
    s = s[0...idx] + '9' + s[idx + 1..-1]
  end
  s.to_i
end

Complexities

• Time: O(n) – n: number of digits
• Space: O(1)