Leetcode #1092: Shortest Common Supersequence
In this guide, we solve Leetcode #1092 **Shortest Common Supersequence ** in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences. If there are multiple valid strings, return any of them.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation:
str1 = "abac" is a subsequence of "cabac" because we can delete the first "c".
str2 = "cab" is a subsequence of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties.
Python Solution
class Solution:
def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
m, n = len(str1), len(str2)
f = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if str1[i - 1] == str2[j - 1]:
f[i][j] = f[i - 1][j - 1] + 1
else:
f[i][j] = max(f[i - 1][j], f[i][j - 1])
ans = []
i, j = m, n
while i or j:
if i == 0:
j -= 1
ans.append(str2[j])
elif j == 0:
i -= 1
ans.append(str1[i])
else:
if f[i][j] == f[i - 1][j]:
i -= 1
ans.append(str1[i])
elif f[i][j] == f[i][j - 1]:
j -= 1
ans.append(str2[j])
else:
i, j = i - 1, j - 1
ans.append(str1[i])
return ''.join(ans[::-1])
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.