LeetCode 1426: Counting Elements Solution

Master LeetCode problem 1426 (Counting Elements), a easy 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.

1426. Counting Elements

Easy

Array Hash Table

Problem Explanation

Explanation:

This problem asks us to find the number of elements in an array that are present in the array but have a count of 'count+1'. We can achieve this by counting the frequency of each element in the array and then checking if the count of 'element+1' is greater than 0.

Algorithmic Steps:

Create a HashMap to store the frequency of each element in the array.
Iterate through the array and populate the HashMap with the frequency of each element.
Iterate through the HashMap, and for each key 'num', check if 'num+1' exists in the HashMap. If it does, increment the result by the frequency of 'num'.

Time Complexity: O(n), where n is the number of elements in the array. Space Complexity: O(n) for the HashMap.

Solution Code

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

class Solution {
    public int countElements(int[] arr) {
        Map<Integer, Integer> freqMap = new HashMap<>();
        int result = 0;

        for (int num : arr) {
            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        }

        for (int num : freqMap.keySet()) {
            if (freqMap.containsKey(num + 1)) {
                result += freqMap.get(num);
            }
        }

        return result;
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 1426 (Counting Elements)?

The time complexity for LeetCode 1426 (Counting Elements) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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