D
DevExCode

LeetCode 2613: Beautiful Pairs Solution

Master LeetCode problem 2613 (Beautiful Pairs), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

The problem asks us to find the maximum number of beautiful pairs we can form by changing exactly one element in one of the arrays. A beautiful pair is a pair of elements (a, b) where a is from the first array and b is from the second array, such that a = b.

To solve this problem, we can count the number of common elements between the two arrays. If the count is equal to the length of the arrays, it means we can change one element to form a beautiful pair. Otherwise, we can change any non-matching element to increase the count of beautiful pairs.

Algorithm:

  1. Initialize a hashmap to store the frequency of elements in the first array.
  2. Iterate through the first array and count the frequency of each element.
  3. Initialize a variable commonCount to count the number of common elements between the two arrays.
  4. Iterate through the second array and for each element:
    • If the element exists in the hashmap and its frequency is greater than 0, increment commonCount and decrement the frequency in the hashmap.
  5. Calculate the maximum number of beautiful pairs maxPairs:
    • If commonCount < length of arrays, maxPairs = commonCount + 1.
    • If commonCount = length of arrays, maxPairs = commonCount - 1.
  6. Return maxPairs.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the length of the arrays.

Space Complexity:

The space complexity of this algorithm is O(n) to store the frequency of elements in the first array.

:

Solution Code

class Solution {
    public int beautifulPairs(int[] A, int[] B) {
        Map<Integer, Integer> freqMap = new HashMap<>();
        for (int num : A) {
            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        }
        
        int commonCount = 0;
        for (int num : B) {
            if (freqMap.containsKey(num) && freqMap.get(num) > 0) {
                commonCount++;
                freqMap.put(num, freqMap.get(num) - 1);
            }
        }
        
        int maxPairs = (commonCount == A.length) ? commonCount - 1 : commonCount + 1;
        return maxPairs;
    }
}

Try It Yourself

Loading code editor...

Related LeetCode Problems

Frequently Asked Questions

How to solve LeetCode 2613 (Beautiful Pairs)?

This page provides optimized solutions for LeetCode problem 2613 (Beautiful Pairs) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 2613 (Beautiful Pairs)?

The time complexity for LeetCode 2613 (Beautiful Pairs) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 2613 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 2613 in Java, C++, or Python.

Back to LeetCode Solutions