leetcode - 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.
1
2
3

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.
1
2
3

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.
1
2
3

Constraints:

1 <= k <= num.length <= 10^5
num consists of only digits.
num does not have any leading zeros except for the zero itself.
1
2
3

Solution

Solved after hints from the tag…

Use a monotonic stack, when the stack is increasing the number is the minimum. Remember when returning the result, delete the leading 0s.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        stack = []
        num = list(num)
        i = 0
        while i < len(num) and k > 0:
            if not stack or stack[-1] <= num[i]:
                stack.append(num[i])
                i += 1
            else:
                stack.pop()
                k -= 1
        if i < len(num):
            stack += num[i:]
        elif k > 0:
            stack = stack[:-k]
        # delete the leading 0
        return ''.join(stack).lstrip('0') or '0'
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18