Count Pairs of Equal Substrings With Minimum Difference Python/Java solution

The typical implementation builds a dictionary or HashMap from characters to index lists in the second string. Then iterate through the first string and compare positions using pointer movement to update the minimum difference and count. The same logic works in Python, Java, C++, Go, and TypeScript with O(n + m) complexity.

What is the best approach for Count Pairs of Equal Substrings With Minimum Difference?

The most efficient solution uses a Greedy + Hash Table strategy. Store all indices of each character from the second string in a hash table, then scan the first string and compare positions using a two‑pointer technique. This reduces unnecessary comparisons and achieves O(n + m) time complexity.

Is Count Pairs of Equal Substrings With Minimum Difference asked at Google/Amazon/Meta?

Problems involving minimum index differences, hash tables, and efficient string matching patterns are common in interviews at companies like Amazon, Google, and Meta. Variants often appear where candidates must optimize brute force substring or character comparisons using hashing or pointer techniques.

What data structure is used in Count Pairs of Equal Substrings With Minimum Difference?

A hash table (or dictionary) is the main data structure. It maps each character to a list of its indices in one of the strings. These index lists are then processed with a greedy two‑pointer technique to compute minimum differences efficiently.

What is the time complexity of Count Pairs of Equal Substrings With Minimum Difference?

The optimal approach runs in O(n + m) time where n and m are the lengths of the two strings. Building the hash table of character positions takes O(m), and scanning the first string with pointer comparisons takes O(n). Space complexity is O(n + m) for storing indices.

Description Hints 2 Solution Video 4:23

1794. Count Pairs of Equal Substrings With Minimum Difference

Q: Is Count Pairs of Equal Substrings With Minimum Difference easy or hard?

This problem is generally classified as Medium. The brute force idea is simple, but finding the optimal solution requires recognizing that characters can be grouped with a hash table and compared using a greedy pointer technique to avoid O(n*m) comparisons.

Q: How to solve Count Pairs of Equal Substrings With Minimum Difference in O(n)?

Group indices of characters from the second string using a hash table. For each character in the first string, access the corresponding index list and compare nearby positions using a two‑pointer scan. Since each pointer moves forward at most once per element, the total work across both strings stays linear.

Medium62.7% acceptance1 companyPremium elsewhere, free here

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 i^th character and ends at the j^th character (inclusive) is equal to the substring of secondString that starts at the a^th character and ends at the b^th character (inclusive).
j - a is the minimum possible value among all quadruples that satisfy the previous conditions.

Return the number of such quadruples.

Example 1:

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.

Example 2:

Input: firstString = "ab", secondString = "cd"
Output: 0
Explanation: There are no quadruples satisfying all the conditions.

Constraints:

1 <= firstString.length, secondString.length <= 2 * 10⁵
Both strings consist only of lowercase English letters.

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

If the chosen substrings are of size larger than 1, then you can remove all but the first character from both substrings, and you'll get equal substrings of size 1, with the same a but less j. Hence, it's always optimal to choose substrings of size 1.

Hint 2

If you choose a specific letter, then it's optimal to choose its first occurrence in firstString, and its last occurrence in secondString, to minimize j-a.

4:23

Still stuck?Read the written solution, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

"abcd"
"bccda"

Input

"ab"
"cd"

Your code is saved on this device.Solution