Skip to content

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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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'

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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"
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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';
}

Comments