How do you solve Count Nice Pairs in an Array on LeetCode?

Count Nice Pairs in an Array (LeetCode #1814) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach checks every possible pair of indices to see if they satisfy the nice pair condition. This is straightforward but inefficient for large arrays.

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

The optimal solution for Count Nice Pairs in an Array runs in O(n²) time and uses O(1) space. This complexity arises because we are checking every pair of indices in a nested loop, leading to n(n-1)/2 comparisons.

What algorithm or data structure is used to solve Count Nice Pairs in an Array?

Count Nice Pairs in an Array uses the following concepts: Array, Hash Table, Math, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Nice Pairs in an Array?

Think about how to reduce the number of comparisons Use HashMap for counting frequencies Always consider edge cases, like single-element arrays

What are common mistakes when solving Count Nice Pairs in an Array?

Not reversing numbers correctly Forgetting to use modulo operation

Count Nice Pairs in an Array — LeetCode #1814 (Medium)

Tags: Array, Hash Table, Math, Counting

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach checks every possible pair of indices to see if they satisfy the nice pair condition. This is straightforward but inefficient for large arrays.

This complexity arises because we are checking every pair of indices in a nested loop, leading to n(n-1)/2 comparisons.

Step 1: Initialize a counter to zero for counting nice pairs.
Step 2: Loop through each pair of indices (i, j) where i < j.
Step 3: For each pair, calculate rev(nums[i]) and rev(nums[j]). Check if nums[i] + rev(nums[j]) equals nums[j] + rev(nums[i]). If true, increment the counter.
Step 4: Return the counter modulo 10^9 + 7.

For nums = [42, 11, 1, 97]: 1. i=0, j=1: 42 + rev(11) vs 11 + rev(42) → 42 + 11 vs 11 + 24 (not equal) 2. i=0, j=2: 42 + rev(1) vs 1 + rev(42) → 42 + 1 vs 1 + 24 (not equal) 3. i=0, j=3: 42 + rev(97) vs 97 + rev(42) → 42 + 79 vs 97 + 24 (equal, count=1) 4. i=1, j=2: 11 + rev(1) vs 1 + rev(11) → 11 + 1 vs 1 + 11 (equal, count=2) 5. i=1, j=3: 11 + rev(97) vs 97 + rev(11) → 11 + 79 vs 97 + 24 (not equal) 6. i=2, j=3: 1 + rev(97) vs 97 + rev(1) → 1 + 79 vs 97 + 1 (not equal) Final count = 2.

Common Mistakes

Not reversing numbers correctly
Forgetting to use modulo operation

Interview Tips

Think about how to reduce the number of comparisons
Use HashMap for counting frequencies
Always consider edge cases, like single-element arrays

#1814

Count Nice Pairs in an Array

Medium

Array↗Hash Table↗Math↗Counting↗Hash Map↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach checks every possible pair of indices to see if they satisfy the nice pair condition. This is straightforward but inefficient for large arrays.

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter to zero for counting nice pairs.
2Step 2: Loop through each pair of indices (i, j) where i < j.
3Step 3: For each pair, calculate rev(nums[i]) and rev(nums[j]). Check if nums[i] + rev(nums[j]) equals nums[j] + rev(nums[i]). If true, increment the counter.
4Step 4: Return the counter modulo 10^9 + 7.

solution.py14 lines

1# Full working Python code
2
3def rev(x):
4    return int(str(x)[::-1])
5
6def countNicePairs(nums):
7    MOD = 10**9 + 7
8    count = 0
9    n = len(nums)
10    for i in range(n):
11        for j in range(i + 1, n):
12            if nums[i] + rev(nums[j]) == nums[j] + rev(nums[i]):
13                count += 1
14    return count % MOD

ℹ

Complexity note: This complexity arises because we are checking every pair of indices in a nested loop, leading to n(n-1)/2 comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.