#3265

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Count Almost Equal Pairs I

Count Almost Equal Pairs I involves finding pairs of elements that can be made equal by swapping at most one digit.

array hash table sorting counting enumeration

Problem Statement

You are given an array of positive integers, nums. Two integers x and y are almost equal if, by swapping at most one digit in either number, you can make them identical. Your task is to return the number of index pairs i and j (i < j) such that nums[i] and nums[j] are almost equal.

To solve this problem, you need to scan through the array, identify pairs of elements that are almost equal by the described operation, and count them. Efficient lookups using hash tables will be beneficial for optimizing the approach, especially when checking multiple pairs within the constraints.

Examples

Example 1

Input: nums = [3,12,30,17,21]

Output: 2

The almost equal pairs of elements are:

Example 2

Input: nums = [1,1,1,1,1]

Output: 10

Every two elements in the array are almost equal.

Example 3

Input: nums = [123,231]

Output: 0

We cannot swap any two digits of 123 or 231 to reach the other.

Constraints

2 <= nums.length <= 100
1 <= nums[i] <= 106

Solution Approach

Brute-force Pair Scanning

A straightforward approach is to check every pair of elements in the array, applying the swap operation to see if the two numbers can be made equal. This is feasible for smaller arrays but may become inefficient for larger arrays due to its O(n^2) complexity.

Hash Table Lookup for Efficiency

Instead of checking all pairs directly, store transformed versions of each number in a hash table. This allows quick lookups to determine if a pair can be made equal after a digit swap, optimizing the process from a brute-force check to a more efficient approach.

Sorting and Grouping

Another approach involves sorting the array and grouping numbers based on their digit structure. This can help identify possible almost equal pairs by looking for numbers that are similar but differ by a single digit.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used. The brute-force approach has a time complexity of O(n^2) due to checking all pairs. The hash table approach can reduce it to O(n) for insertion and lookup but requires handling the number transformation. Sorting-based approaches can range from O(n log n) to O(n^2) depending on how efficiently the pairs are identified post-sorting.

What Interviewers Usually Probe

Check for familiarity with array scanning techniques.
Assess ability to optimize brute-force solutions with hash tables.
Evaluate understanding of sorting and its impact on time complexity.

Common Pitfalls or Variants

Common pitfalls

Assuming a brute-force approach is always efficient for arrays of any size.
Neglecting to account for edge cases where numbers cannot be made equal with a single swap.
Overcomplicating the solution by applying complex sorting methods where hash table lookups suffice.

Follow-up variants

Allow for a maximum number of allowed swaps (not just one).
Implement a solution that can handle non-digit swaps or multiple digits being swapped.
Optimize for larger arrays, introducing additional constraints on time or space.

FAQ

What is the best approach for solving Count Almost Equal Pairs I?

The optimal approach uses array scanning combined with hash table lookups to efficiently check for almost equal pairs.

How can I optimize a brute-force solution for this problem?

By using a hash table to store and lookup possible transformed versions of each number, you can reduce the time complexity significantly.

What is the time complexity of the brute-force approach in Count Almost Equal Pairs I?

The brute-force approach has a time complexity of O(n^2) as it involves checking all pairs in the array.

Can sorting the array help with solving this problem?

Yes, sorting the array can group similar numbers together, which might help identify almost equal pairs more efficiently, though this approach has a complexity of O(n log n).

What edge cases should I consider for Count Almost Equal Pairs I?

Consider cases where no two numbers are almost equal, as well as cases where all numbers are the same or differ by more than one digit.

terminal

Solution

Solution 1: Sorting + Enumeration

We can enumerate each number, and for each number, we can enumerate each pair of different digits, then swap these two digits to get a new number. We record this new number in a hash table $s$, representing all possible numbers after at most one swap. Then, we count how many numbers previously enumerated are in the hash table $s$ and add this count to the answer. Next, we add the currently enumerated number to the hash table $\textit{cnt}$, representing the count of the current number.

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class Solution:
    def countPairs(self, nums: List[int]) -> int:
        nums.sort()
        ans = 0
        cnt = defaultdict(int)
        for x in nums:
            vis = {x}
            s = list(str(x))
            for j in range(len(s)):
                for i in range(j):
                    s[i], s[j] = s[j], s[i]
                    vis.add(int("".join(s)))
                    s[i], s[j] = s[j], s[i]
            ans += sum(cnt[x] for x in vis)
            cnt[x] += 1
        return ans

Continue Practicing

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