Remove K Digits

Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/remove-k-digits ## Basic Ideas: Greedy ## From left to right, if current digit decreases, we find one candiate ## Why? ## i-1, i, i +1 ## digit[i-1] <= digit[i] and digit[i] > digit[i+1] ## Let's say we don't delete digit i. ## If the final result has deleted digit i+1, the result won't be optimal. ## Since we can keep digit i+1 and delete digit i, with everything else unchanged. ## The modified result would be smaller ## If the final result hasn't deleted digit i+1 ## If the deletion has happened before i, it doesn't make sense ## If the deletion has happened after i+1. ## We can find another combination which is better. ## Delete digit i and keep any other deletion. ## ## Complexity: Time O(n), Space O(1) class Solution: def removeKdigits(self, num, k): """ :type num: str :type k: int :rtype: str """ length = len(num) l = [] for i in range(length): # delete the previous, if necessary while len(l) != 0 and k > 0: if num[i] < l[-1]: del l[-1] k -= 1 else: break if k == 0: l.append(num[i]) else: if i == length -1 or num[i] <= num[i+1]: l.append(num[i]) else: # should delete digit i k -= 1 # remove from the tail while len(l) !=0 and k>0: del l[-1] k -= 1 res = ''.join(l).lstrip('0') if res == '': res ='0' return res # s = Solution() # print(s.removeKdigits("12", 1))

Share It, If You Like It.