402. Remove K Digits

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.

Approach #1: C++. [brute force]

class Solution {
public:
    string removeKdigits(string num, int k) {
        int len = num.length();
        if (len == k) return "0";
        int rmd = len - k;
        string ans = "";
        int idx = 0;
        for (int i = 0; i < len; i = idx+1) {
            int temp = INT_MAX;
            rmd--;
            for (int j = i; j < len-rmd; ++j) {
                if (num[j]-'0' < temp) {
                    temp = num[j] - '0';
                    idx = j;
                }
            }
            ans += to_string(temp);
            if (ans.length() == len - k) break;
        }
        while (ans[0] == '0' && ans.length() > 1) 
            ans = ans.substr(1);
        return ans;
    }
};

　　

Time complex: O(n^2)

Approach #2:  Java. [stack]

class Solution {
    public String removeKdigits(String num, int k) {
        int digits = num.length() - k;
        char[] stk = new char[num.length()];
        int top = 0;
        for (int i = 0; i < num.length(); ++i) {
            char c = num.charAt(i);
            while (top > 0 && stk[top-1] > c && k > 0) {
                top--;
                k--;
            }
            stk[top++] = c;
        }
        int idx = 0;
        while (idx < digits && stk[idx] == '0') idx++;
        return idx == digits ? "0" : new String(stk, idx, digits-idx);
    }
}

　　

Time complex: O(n).