EasyNo. 2937
Make Three Strings Equal — LeetCode 2937 Python Solution
- Pattern
- Hash Map
- Topics
- String
- Reading time
- 2 min
- Source
- leetcode.com
The problem
You are given three strings: s1, s2, and s3. In one operation you can choose one of these strings and delete its rightmost character.
Python solution
class Solution:
def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
s = len(s1) + len(s2) + len(s3)
n = min(len(s1), len(s2), len(s3))
for i in range(n):
if not s1[i] == s2[i] == s3[i]:
return -1 if i == 0 else s - 3 * i
return s - 3 * nComplexity
- Time
- O(n)where n is the minimum length of the three strings
- Space · auxiliary
- O(1)
Pattern: Hash Map
Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 2937. Make Three Strings Equal is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.
The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.
Related problems
| No. | Problem | Difficulty |
|---|---|---|
| 2942 | Find Words Containing Character | Easy |
| 0006 | Zigzag Conversion | Medium |
| 0008 | String to Integer (atoi) | Medium |
| 0038 | Count and Say | Medium |
| 0058 | Length of Last Word | Easy |
| 0065 | Valid Number | Hard |
Frequently asked questions
- How hard is LeetCode 2937. Make Three Strings Equal?
LeetCode 2937. Make Three Strings Equal is rated Easy on LeetCode.
- What is the time complexity of LeetCode 2937. Make Three Strings Equal?
The Python solution on this page runs in O(n), where n is the minimum length of the three strings.
- What is the space complexity of LeetCode 2937. Make Three Strings Equal?
The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 2937. Make Three Strings Equal cover?
LeetCode 2937. Make Three Strings Equal is tagged String on LeetCode.