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;
}
};
