402. Remove K Digits

https://leetcode.com/problems/remove-k-digits/

Problem

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible.

Note:

• The length of num is less than 10002 and will be ≥ k.

• The given num does not contain any leading zero.

Example 1:

Input: num = "1432219", k = 3
Output: "1219"
Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.

Example 2:

Input: num = "10200", k = 1
Output: "200"
Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.

Example 3:

Input: num = "10", k = 2
Output: "0"
Explanation: Remove all the digits from the number and it is left with nothing which is 0.

Solution

1. Naive (recursive)

class Solution {
public:
    string removeKdigits(string &num, int k) {
        if (k == 0) return num;
        const int n = num.size();
        if (k >= n) return "0";
        
        int i = 0;
        for (; i < n - 1; ++i) {
            if (num[i] > num[i + 1]) break;
        }
        
        int j = i;
        if (i == 0) while (j + 1 < n && num[j + 1] == '0') ++j;
        string str = num.substr(0, i) + num.substr(j + 1, n - j);
        str = (str.empty()) ? "0" : str;
        return removeKdigits(str, k - 1);
    }
};

2. Stack

class Solution {
public:
    string removeKdigits(string &num, int k) {
        string res;
        for (const auto c: num) {
            while (!res.empty() && res.back() > c && k-- > 0) res.pop_back();
            res.push_back(c);
        }
        while (k-- > 0) res.pop_back();
        int i = 0;
        while (i < res.size() && res[i] == '0') ++i;
        res.erase(res.begin(), res.begin() + i);
        return (res.empty() || (int) res.size() <= k) ? "0" : res;
    }
};

• #string

• #recursive

• #stack