What is the time complexity of Count Almost Equal Pairs I?

The optimal solution has a time complexity of O(n^2 * log(max_num)).

What is the space complexity of Count Almost Equal Pairs I?

The space complexity is O(log(max_num)).

What data structures are used to solve Count Almost Equal Pairs I?

This problem uses Array, Hash Table, Sorting, Counting, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Almost Equal Pairs I?

Insight 1: The core is to implement a function that checks if two numbers are 'almost equal' efficiently. Insight 2: Converting numbers to strings makes digit manipulation and comparison easier.

Which companies ask LeetCode 3265. Count Almost Equal Pairs I in interviews?

This problem is frequently asked at Info Edge.

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Count Almost Equal Pairs I - LeetCode 3265 Solution

Q: How to solve Count Almost Equal Pairs I?

The solution iterates through all possible pairs (i, j) in the input array `nums` where i < j. For each pair, it calls a helper function `are_almost_equal` to determine if the two numbers at those indices are almost equal. The `are_almost_equal` function converts the integers to strings, pads them with leading zeros if necessary to ensure equal length, and then compares them character by character. If the strings are identical, they are almost equal. If there are exactly two different characters, it checks if swapping them would make the strings identical. If so, the numbers are almost equal. Finally, the outer loop increments a counter whenever an almost equal pair is found, returning the final count.

Count Almost Equal Pairs I - Complete Solution Guide

Count Almost Equal Pairs I is LeetCode problem 3265, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array nums consisting of positive integers. We call two integers x and y in this problem almost equal if both integers can become equal after performing the following operation at most once : Choose either x or y and swap any two digits within the chosen number. Return the number of indices i and j in nums where i < j such that nums[i] and nums[j] are almost equal . Note that it is allowed for an integer to have leading zeros after performing an operation. Example 1: Input: nums

Detailed Explanation

The problem requires us to count the number of almost equal pairs in a given array of positive integers. Two integers are considered 'almost equal' if they are equal, or can become equal by swapping two digits within either of the numbers at most once. The comparison should only be done between pairs of numbers with indices i and j where i < j. The numbers can have leading zeros after the swap.

Solution Approach

The solution iterates through all possible pairs (i, j) in the input array `nums` where i < j. For each pair, it calls a helper function `are_almost_equal` to determine if the two numbers at those indices are almost equal. The `are_almost_equal` function converts the integers to strings, pads them with leading zeros if necessary to ensure equal length, and then compares them character by character. If the strings are identical, they are almost equal. If there are exactly two different characters, it checks if swapping them would make the strings identical. If so, the numbers are almost equal. Finally, the outer loop increments a counter whenever an almost equal pair is found, returning the final count.

Step-by-Step Algorithm

Step 1: Iterate through the `nums` array with nested loops to consider all pairs (i, j) such that i < j.
Step 2: For each pair (nums[i], nums[j]), call the `are_almost_equal` function.
Step 3: Inside `are_almost_equal`, convert both integers to strings.
Step 4: Determine the maximum length of the two strings.
Step 5: Pad both strings with leading zeros until their lengths match the maximum length.
Step 6: If the padded strings are identical, return `true`.
Step 7: If the strings are not identical, identify the indices where the characters differ.
Step 8: If the number of differing indices is exactly two, check if swapping the characters at those indices in one of the strings would make the strings identical.
Step 9: If the swap would make the strings identical, return `true`.
Step 10: If any of the conditions for almost equality are not met, return `false`.
Step 11: In the outer loop, if `are_almost_equal` returns `true`, increment the counter.
Step 12: After iterating through all pairs, return the final count.

Key Insights

Insight 1: The core is to implement a function that checks if two numbers are 'almost equal' efficiently.
Insight 2: Converting numbers to strings makes digit manipulation and comparison easier.
Insight 3: Padding strings with leading zeros simplifies the comparison when the numbers have different lengths.
Insight 4: The number of different digits should be checked whether it is equal to zero or two.

Complexity Analysis

Time Complexity: O(n^2 * log(max_num))

Space Complexity: O(log(max_num))

Topics

This problem involves: Array, Hash Table, Sorting, Counting, Enumeration.

Companies

Asked at: Info Edge.

Count Almost Equal Pairs I | LeetCode 3265

Medium

Array Hash Table Sorting Counting Enumeration

Time: O(n^2 * log(max_num))Space: O(log(max_num))

Solution Code

Available in:

class Solution {
    public int countPairs(int[] nums) {
        
        class Helper {
            public boolean areAlmostEqual(int x, int y) {
                String s_x = String.valueOf(x);
                String s_y = String.valueOf(y);

                int max_len = Math.max(s_x.length(), s_y.length());

                s_x = String.format("%" + max_len + "s", s_x).replace(' ', '0');
                s_y = String.format("%" + max_len + "s", s_y).replace(' ', '0');

                if (s_x.equals(s_y)) {
                    return true;
                }

                java.util.List<Integer> diff_indices = new java.util.ArrayList<>();
                for (int i = 0; i < max_len; i++) {
                    if (s_x.charAt(i) != s_y.charAt(i)) {
                        diff_indices.add(i);
                    }
                }

                if (diff_indices.size() == 2) {
                    int i = diff_indices.get(0);
                    int j = diff_indices.get(1);
                    if (s_x.charAt(i) == s_y.charAt(j) && s_x.charAt(j) == s_y.charAt(i)) {
                        return true;
                    }
                }

                return false;
            }
        }

        Helper helper = new Helper();
        int count = 0;
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (helper.areAlmostEqual(nums[i], nums[j])) {
                    count++;
                }
            }
        }

        return count;
    }
}

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Problem Info

Description

Problem 3265: Count Almost Equal Pairs I - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Counting Enumeration

Companies

Info Edge

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3265

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Iterate through the `nums` array with nested loops to consider all pairs (i, j) such that i < j.
2
Step 2: For each pair (nums[i], nums[j]), call the `are_almost_equal` function.
3
Step 3: Inside `are_almost_equal`, convert both integers to strings.
4
Step 4: Determine the maximum length of the two strings.
5
Step 5: Pad both strings with leading zeros until their lengths match the maximum length.
6
Step 6: If the padded strings are identical, return `true`.
7
Step 7: If the strings are not identical, identify the indices where the characters differ.
8
Step 8: If the number of differing indices is exactly two, check if swapping the characters at those indices in one of the strings would make the strings identical.
9
Step 9: If the swap would make the strings identical, return `true`.
10
Step 10: If any of the conditions for almost equality are not met, return `false`.
11
Step 11: In the outer loop, if `are_almost_equal` returns `true`, increment the counter.
12
Step 12: After iterating through all pairs, return the final count.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2 * log(max_num))

💾

Space Complexity

O(log(max_num))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is dominated by the nested loops which iterate through all pairs O(n^2). The string operations inside the helper function contribute O(log(max_num)) to each iteration. The space complexity arises from storing the numbers as strings.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting and Hashing

An alternative approach would be to sort the digits of each number and store the sorted strings in a hash table. This would allow for efficient lookup of numbers that are almost equal because they have the same sorted digits. After hashing, count all the same elements in the table, then iterate through the original list comparing element i and j, if they differ by only one swap after being sorted.

Time: O(n * klogk) for sorting the digits of each number and O(n^2*k) to check swap, where k is the maximum number of digits in a number.Space: O(n), due to the space used by hash table.

Trade-offs: The initial sorting step is slower than simply comparing the numbers directly. However, the hashing allows for faster duplicate finding among almost equal numbers. Can be better if there are many duplicates after the sorting step.

✨

Key Insights

1Insight 1: The core is to implement a function that checks if two numbers are 'almost equal' efficiently.
2Insight 2: Converting numbers to strings makes digit manipulation and comparison easier.
3Insight 3: Padding strings with leading zeros simplifies the comparison when the numbers have different lengths.
4Insight 4: The number of different digits should be checked whether it is equal to zero or two.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `nums = [1, 1, 1, 1, 1]`. Every pair is almost equal. The code correctly counts all pairs.
Edge case 2: `nums = [1, 10]`. The code pads 1 to '01' and 10 remains '10'. Since s_x[0] == s_y[1] == '1' and s_x[1] == s_y[0] == '0', then those number are considered almost equal
Edge case 3: Numbers with leading zeros like `nums = [01, 10]` or `nums = [001, 100]` should be considered. Code correctly handles leading zeros.
Edge case 4: Numbers that are equal like nums = [12, 12]. The function areAlmostEqual will identify it by returning true since no swap is required.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling leading zeros correctly. Numbers like 1 and 01 should be considered almost equal, and failing to pad them with zeros can lead to incorrect results.

✗Common mistake 2: Incorrectly implementing the 'almost equal' check. Forgetting the case where two numbers are already equal, or not correctly checking for a valid swap.

✗Common mistake 3: Not considering all possible pairs (i < j) leading to missed pairs.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data cleaning and deduplication. Identifying nearly identical entries in a database that may have typos or slightly different formatting.

🏦Real-world scenario 2: Fuzzy matching in search engines. Allowing users to find results even if they misspell a search query or enter slightly incorrect data.

🎮Real-world scenario 3: Code plagiarism detection. Identifying code snippets that are very similar, even if variable names or minor changes have been made.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Almost Equal Pairs I"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?