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Count Pairs of Equal Substrings With Minimum Difference

Medium
25%
Updated 8/1/2025

Asked by 1 Company

Count Pairs of Equal Substrings With Minimum Difference

What is this problem about?

The "Count Pairs of Equal Substrings With Minimum Difference interview question" is a string alignment challenge. You are given two strings, s1s1 and s2s2. You need to find pairs of indices (i,j)(i, j) such that the character s1[i]s1[i] is equal to s2[j]s2[j]. Among all such pairs, you are specifically interested in those where the difference between the indices (ij)(i - j) is minimized. Your goal is to count how many such "minimal difference" pairs exist.

Why is this asked in interviews?

Google uses the "Count Pairs of Equal Substrings With Minimum Difference coding problem" to test a candidate's ability to optimize search operations using hash tables. It evaluates if you can identify that comparing every character in s1s1 with every character in s2s2 (O(NimesM)O(N imes M)) is unnecessary. It’s a test of "Hash Table interview pattern" and "Greedy" logic.

Algorithmic pattern used

This problem relies on Hash Tables and Prefix/Suffix Tracking.

  1. First Occurrences: For string s1s1, we want the first occurrence of each character to minimize ii. Store these in a map: first_pos[char] = index.
  2. Last Occurrences: For string s2s2, we want the last occurrence of each character to maximize jj, which in turn minimizes the difference iji - j. Store these in last_pos[char] = index.
  3. Minimize Difference: Iterate through all characters 'a' through 'z'. If a character exists in both maps, calculate diff = first_pos[char] - last_pos[char].
  4. Count: Keep track of the global minimum diff found so far and count how many characters achieve it.

Example explanation

s1="abacaba"s1 = "abacaba", s2="abacaba"s2 = "abacaba"

  • First occurrences in s1s1: {a:0, b:1, c:3}
  • Last occurrences in s2s2: {a:6, b:5, c:3}
  • Differences:
    • 'a': 06=60 - 6 = -6
    • 'b': 15=41 - 5 = -4
    • 'c': 33=03 - 3 = 0 The minimum difference is -6 (for 'a'). There is only 1 such pair.

Common mistakes candidates make

  • Brute Force: Checking all NimesMN imes M pairs of indices.
  • Incorrect logic for "Minimum": Trying to minimize ij|i - j| instead of iji - j. The problem usually specifies the direction or specific index subtraction order.
  • Ignoring characters: Forgetting that only equal characters can form a pair.

Interview preparation tip

When asked to find a "minimum difference" in a string or array, always consider if pre-calculating the first or last occurrence of elements helps. This "Greedy interview pattern" is a standard way to avoid nested loops.

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