Description
Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num.
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.
Constraints:
1 <= k <= num.length <= 105num consists of only digits.num does not have any leading zeros except for the zero itself.
Solutions
Solution 1: Greedy Algorithm
PythonJavaC++GoTypeScript
class Solution:
def removeKdigits(self, num: str, k: int) -> str:
stk = []
remain = len(num) - k
for c in num:
while k and stk and stk[-1] > c:
stk.pop()
k -= 1
stk.append(c)
return ''.join(stk[:remain]).lstrip('0') or '0'(code-box)
class Solution {
public String removeKdigits(String num, int k) {
StringBuilder stk = new StringBuilder();
for (char c : num.toCharArray()) {
while (k > 0 && stk.length() > 0 && stk.charAt(stk.length() - 1) > c) {
stk.deleteCharAt(stk.length() - 1);
--k;
}
stk.append(c);
}
for (; k > 0; --k) {
stk.deleteCharAt(stk.length() - 1);
}
int i = 0;
for (; i < stk.length() && stk.charAt(i) == '0'; ++i) {
}
String ans = stk.substring(i);
return "".equals(ans) ? "0" : ans;
}
}(code-box)
class Solution {
public:
string removeKdigits(string num, int k) {
string stk;
for (char& c : num) {
while (k && stk.size() && stk.back() > c) {
stk.pop_back();
--k;
}
stk += c;
}
while (k--) {
stk.pop_back();
}
int i = 0;
for (; i < stk.size() && stk[i] == '0'; ++i) {
}
string ans = stk.substr(i);
return ans == "" ? "0" : ans;
}
};(code-box)
func removeKdigits(num string, k int) string {
stk, remain := make([]byte, 0), len(num)-k
for i := 0; i < len(num); i++ {
n := len(stk)
for k > 0 && n > 0 && stk[n-1] > num[i] {
stk = stk[:n-1]
n, k = n-1, k-1
}
stk = append(stk, num[i])
}
for i := 0; i < len(stk) && i < remain; i++ {
if stk[i] != '0' {
return string(stk[i:remain])
}
}
return "0"
}(code-box)
function removeKdigits(num: string, k: number): string {
const stk: string[] = [];
for (const c of num) {
while (k && stk.length > 0 && stk[stk.length - 1] > c) {
stk.pop();
k--;
}
stk.push(c);
}
while (k--) {
stk.pop();
}
return stk.join('').replace(/^0*/g, '') || '0';
}(code-box)
