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LeetCode 1296: Divide Array in Sets of K Consecutive Numbers Solution

Master LeetCode problem 1296 (Divide Array in Sets of K Consecutive Numbers), a medium 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.

1296. Divide Array in Sets of K Consecutive Numbers

Medium
ArrayHash TableGreedySorting

Problem Explanation

Explanation:

To solve this problem, we can use a hashmap to store the frequencies of each number in the input array nums. We then iterate through the keys of the hashmap in ascending order, and for each key, we check if there are enough occurrences of consecutive numbers starting from that key to form sets of size k.

  1. Create a hashmap freqMap to store the frequencies of each number in nums.
  2. Sort the keys of freqMap.
  3. For each key in the sorted keys:
    • If the frequency of the current key in freqMap is greater than 0:
      • Check if there are enough occurrences of consecutive numbers starting from the current key to form sets of size k.
      • If yes, reduce the frequencies of those consecutive numbers in freqMap.
      • If no, return false.
  4. If we successfully iterate through all keys, return true.

Time Complexity: O(n log n) where n is the size of nums (due to sorting) Space Complexity: O(n) where n is the size of nums

Solution Code

import java.util.*;

class Solution {
    public boolean isPossibleDivide(int[] nums, int k) {
        Map<Integer, Integer> freqMap = new HashMap<>();
        for (int num : nums) {
            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        }
        
        List<Integer> keys = new ArrayList<>(freqMap.keySet());
        Collections.sort(keys);

        for (int key : keys) {
            int freq = freqMap.get(key);
            if (freq > 0) {
                for (int i = 0; i < k; i++) {
                    if (freqMap.getOrDefault(key + i, 0) < freq) {
                        return false;
                    }
                    freqMap.put(key + i, freqMap.get(key + i) - freq);
                }
            }
        }
        
        return true;
    }
}

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