Leetcode #1794: Count Pairs of Equal Substrings With Minimum Difference
In this guide, we solve Leetcode #1794 Count Pairs of Equal Substrings With Minimum Difference 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
You are given two strings firstString and secondString that are 0-indexed and consist only of lowercase English letters. Count the number of index quadruples (i,j,a,b) that satisfy the following conditions: 0 <= i <= j < firstString.length 0 <= a <= b < secondString.length The substring of firstString that starts at the ith character and ends at the jth character (inclusive) is equal to the substring of secondString that starts at the ath character and ends at the bth character (inclusive).
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Greedy, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: firstString = "abcd", secondString = "bccda"
Output: 1
Explanation: The quadruple (0,0,4,4) is the only one that satisfies all the conditions and minimizes j - a.
Python Solution
class Solution:
def countQuadruples(self, firstString: str, secondString: str) -> int:
last = {c: i for i, c in enumerate(secondString)}
ans, mi = 0, inf
for i, c in enumerate(firstString):
if c in last:
t = i - last[c]
if mi > t:
mi = t
ans = 1
elif mi == t:
ans += 1
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.