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738. Monotone Increasing Digits

Description

An integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.

Given an integer n, return the largest number that is less than or equal to n with monotone increasing digits.

 

Example 1:

Input: n = 10
Output: 9

Example 2:

Input: n = 1234
Output: 1234

Example 3:

Input: n = 332
Output: 299

 

Constraints:

  • 0 <= n <= 109

Solutions

Solution 1

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class Solution:
    def monotoneIncreasingDigits(self, n: int) -> int:
        s = list(str(n))
        i = 1
        while i < len(s) and s[i - 1] <= s[i]:
            i += 1
        if i < len(s):
            while i and s[i - 1] > s[i]:
                s[i - 1] = str(int(s[i - 1]) - 1)
                i -= 1
            i += 1
            while i < len(s):
                s[i] = '9'
                i += 1
        return int(''.join(s))

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class Solution {
    public int monotoneIncreasingDigits(int n) {
        char[] s = String.valueOf(n).toCharArray();
        int i = 1;
        for (; i < s.length && s[i - 1] <= s[i]; ++i)
            ;
        if (i < s.length) {
            for (; i > 0 && s[i - 1] > s[i]; --i) {
                --s[i - 1];
            }
            ++i;
            for (; i < s.length; ++i) {
                s[i] = '9';
            }
        }
        return Integer.parseInt(String.valueOf(s));
    }
}

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class Solution {
public:
    int monotoneIncreasingDigits(int n) {
        string s = to_string(n);
        int i = 1;
        for (; i < s.size() && s[i - 1] <= s[i]; ++i)
            ;
        if (i < s.size()) {
            for (; i > 0 && s[i - 1] > s[i]; --i) {
                --s[i - 1];
            }
            ++i;
            for (; i < s.size(); ++i) {
                s[i] = '9';
            }
        }
        return stoi(s);
    }
};

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func monotoneIncreasingDigits(n int) int {
    s := []byte(strconv.Itoa(n))
    i := 1
    for ; i < len(s) && s[i-1] <= s[i]; i++ {
    }
    if i < len(s) {
        for ; i > 0 && s[i-1] > s[i]; i-- {
            s[i-1]--
        }
        i++
        for ; i < len(s); i++ {
            s[i] = '9'
        }
    }
    ans, _ := strconv.Atoi(string(s))
    return ans
}

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