# 402. Remove K Digits

## 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 <= 105
• num consists of only digits.
• num does not have any leading zeros except for the zero itself.

## Solutions

Solution 1: Greedy Algorithm

• 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;
}
}

• 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;
}
};

• 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'


• 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"
}

• function removeKdigits(num: string, k: number): string {
let nums = [...num];
while (k > 0) {
let idx = 0;
while (idx < nums.length - 1 && nums[idx + 1] >= nums[idx]) {
idx++;
}
nums.splice(idx, 1);
k--;
}
return nums.join('').replace(/^0*/g, '') || '0';
}