What is the time complexity of Count Nice Pairs in an Array?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Nice Pairs in an Array?

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

What are the key insights for solving Count Nice Pairs in an Array?

Insight 1: The core condition `nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])` can be simplified to `nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])`. This means we only need to calculate this difference for each number and count how many numbers have the same difference. Insight 2: Using a hash table (or dictionary) to store the frequency of each difference is crucial for efficiently counting the pairs. This allows us to achieve a linear time complexity.

Which companies ask LeetCode 1814. Count Nice Pairs in an Array in interviews?

This problem is frequently asked at Block.

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Count Nice Pairs in an Array - LeetCode 1814 Solution

Q: How to solve Count Nice Pairs in an Array?

The solution calculates the difference between each number in the input array and its reverse. It then uses a hash table to store the frequency of each difference encountered. For each number, it checks how many times its calculated difference has already been seen. This count is added to the total number of nice pairs. Finally, the total count is returned modulo 10^9 + 7.

Count Nice Pairs in an Array - Complete Solution Guide

Count Nice Pairs in an Array is LeetCode problem 1814, 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 that consists of non-negative integers. Let us define rev(x) as the reverse of the non-negative integer x . For example, rev(123) = 321 , and rev(120) = 21 . A pair of indices (i, j) is nice if it satisfies all of the following conditions: 0 <= i < j < nums.length nums[i] + rev(nums[j]) == nums[j] + rev(nums[i]) Return the number of nice pairs of indices . Since that number can be too large, return it modulo 10 9 + 7 . Example 1: Input: nums = [42,11,1,97] Output: 2 E

Detailed Explanation

The problem asks us to find the number of 'nice pairs' in a given array of non-negative integers. A pair of indices (i, j) is considered 'nice' if i < j and nums[i] + rev(nums[j]) == nums[j] + rev(nums[i]), where rev(x) is the reverse of the integer x. The final answer should be returned modulo 10^9 + 7.

Solution Approach

The solution calculates the difference between each number in the input array and its reverse. It then uses a hash table to store the frequency of each difference encountered. For each number, it checks how many times its calculated difference has already been seen. This count is added to the total number of nice pairs. Finally, the total count is returned modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize a hash table `freq` to store the frequency of the differences between numbers and their reverses. Also initialize a variable `count` to 0 to keep track of the number of nice pairs.
Step 2: Iterate through the input array `nums`.
Step 3: For each number `num` in `nums`, calculate its reverse `rev_num`.
Step 4: Calculate the difference `diff = num - rev_num`.
Step 5: Check if `diff` exists as a key in the `freq` hash table. If it does, increment `count` by the value associated with `diff` (which is the number of times that difference has appeared so far). Remember to take the modulo 10^9 + 7 of the `count` after incrementing.
Step 6: Increment the frequency of `diff` in the `freq` hash table.
Step 7: After iterating through all the numbers in `nums`, return `count`.

Key Insights

Insight 1: The core condition `nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])` can be simplified to `nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])`. This means we only need to calculate this difference for each number and count how many numbers have the same difference.
Insight 2: Using a hash table (or dictionary) to store the frequency of each difference is crucial for efficiently counting the pairs. This allows us to achieve a linear time complexity.
Insight 3: The modulo operation is necessary to prevent integer overflow, as the number of nice pairs can be quite large.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Math, Counting.

Companies

Asked at: Block.

Count Nice Pairs in an Array | LeetCode 1814

Medium

Array Hash Table Math Counting

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int countNicePairs(int[] nums) {
        int MOD = 1000000007;
        Map<Integer, Integer> freq = new HashMap<>();
        int count = 0;
        
        for (int num : nums) {
            int rev_num = Integer.parseInt(new StringBuilder(String.valueOf(num)).reverse().toString());
            int diff = num - rev_num;
            
            count = (count + freq.getOrDefault(diff, 0)) % MOD;
            freq.put(diff, freq.getOrDefault(diff, 0) + 1);
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 1814: Count Nice Pairs in an Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Counting

Companies

Block

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1814

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

7 steps

1
Step 1: Initialize a hash table `freq` to store the frequency of the differences between numbers and their reverses. Also initialize a variable `count` to 0 to keep track of the number of nice pairs.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each number `num` in `nums`, calculate its reverse `rev_num`.
4
Step 4: Calculate the difference `diff = num - rev_num`.
5
Step 5: Check if `diff` exists as a key in the `freq` hash table. If it does, increment `count` by the value associated with `diff` (which is the number of times that difference has appeared so far). Remember to take the modulo 10^9 + 7 of the `count` after incrementing.
6
Step 6: Increment the frequency of `diff` in the `freq` hash table.
7
Step 7: After iterating through all the numbers in `nums`, return `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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we iterate through the input array once. The space complexity is O(n) because, in the worst case, the hash map can store up to n entries.

📖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

Brute Force

Iterate through all possible pairs (i, j) with i < j and check if nums[i] + rev(nums[j]) == nums[j] + rev(nums[i]).

Time: O(n^2), where n is the length of the input array.Space: O(1)

Trade-offs: The brute force approach is simple to understand but much slower than the hash table approach for larger input arrays. It's only viable for very small input sizes.

✨

Key Insights

1Insight 1: The core condition `nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])` can be simplified to `nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])`. This means we only need to calculate this difference for each number and count how many numbers have the same difference.
2Insight 2: Using a hash table (or dictionary) to store the frequency of each difference is crucial for efficiently counting the pairs. This allows us to achieve a linear time complexity.
3Insight 3: The modulo operation is necessary to prevent integer overflow, as the number of nice pairs can be quite large.

📚

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!

Edge Cases

Edge case 1: Numbers with leading zeros in their reversed form, e.g., reversing 10 yields 1 instead of 01. The given problem statement clarifies how the `rev(x)` function behaves, and the provided solutions correctly implement this behavior.
Edge case 2: Large input arrays. The modulo operation is crucial to prevent integer overflow when counting nice pairs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the modulo operation. The number of nice pairs can be very large, exceeding the maximum integer value, leading to incorrect results without the modulo operation.

✗Common mistake 2: Not simplifying the core equation. Trying to directly compare `nums[i] + rev(nums[j])` with `nums[j] + rev(nums[i])` for all pairs without recognizing the simplification to `nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])` makes the problem harder to solve efficiently.

✗Common mistake 3: Integer overflow when calculating the reversed number. Using appropriate data types (e.g., `long long` in C++) can help avoid this.

✅

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 analysis where finding patterns or relationships between numbers and their reversed counterparts is useful. For example, in financial data analysis, one might look for relationships between stock prices and their reversed representations.

🏦Real-world scenario 2: Cryptography where number manipulation and pattern recognition are important for encoding and decoding messages. The concept of reversing numbers can be a part of a more complex encryption algorithm.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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 15, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Nice Pairs in an Array"?

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.

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