**Solution**

**Remove K Digits** https://leetcode.com/problems/remove-k-digits/

- Examples which are interesting: "1234567", "7654321". The first is strictly increasing and the second is strictly decreasing.
- How do you choose to remove the first element?
- You will scan from left to right and remove the first peak element. Example 1 that would be 7, the last element. Example 2 that would be 7, the first

element. Why the first peak element? You want to remove the first largest element from left which has a smaller element on right, so that the smaller element can fill in for the large element that we removed. You will repeat this process k times and have a complexity of O(K * N). https://discuss.leetcode.com/topic/59871/two-algorithms-with-detailed-explaination - Now how can we do the above efficiently? We can use a stack to simulate this.
- if x > st[-1], push x. If x < st[-1], pop all elements from the stack which are greater than x while making sure we do not pop more than a total of k elements.
- Adjust for boundary conditions.

```
for x in num:
while k and st and x < st[-1]:
st.pop()
k = k-1
st.append(x)
```

**Solution Code**

```
class Solution(object):
def removeKdigits(self, num, k):
"""
:type num: str
:type k: int
:rtype: str
"""
st = []
for x in num:
while k and st and x < st[-1]:
st.pop()
k = k-1
st.append(x)
while st and k > 0:
st.pop()
k = k-1
i = 0
while i < len(st) and st[i] == "0":
i = i+1
if len(st[i:]) == 0:
return "0"
else:
return "".join(st[i:])
```