LeetCode 2369: Check if There is a Valid Partition For The Array Solution

Master LeetCode problem 2369 (Check if There is a Valid Partition For The Array), 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.

2369. Check if There is a Valid Partition For The Array

Medium

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we iterate through the array checking for valid partitions based on the given conditions. We keep track of the current contiguous subarray length and its characteristics (e.g., all elements are the same, 3 consecutive increasing elements). If we encounter a subarray that does not meet any of the conditions, we reset the length and characteristics and continue checking. If at any point we find a valid partition, we return true. If we reach the end of the array without finding a valid partition, we return false.

Time complexity: O(n) where n is the length of the input array nums.
Space complexity: O(1) as we are using constant extra space.

Solution Code

class Solution {
    public boolean checkPartitioning(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n - 2; i++) {
            int a = nums[i], b = nums[i + 1], c = nums[i + 2];
            if ((a == b && b == c) || (a + 1 == b && b + 1 == c) || (a == c)) {
                return true;
            }
        }
        return false;
    }
}

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