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

Problem

You are given a 0-indexed integer array nums. You have to partition the array into one or more contiguous subarrays.

We call a partition of the array valid if each of the obtained subarrays satisfies one of the following conditions:

• The subarray consists of exactly 2 equal elements. For example, the subarray [2,2] is good.

• The subarray consists of exactly 3 equal elements. For example, the subarray [4,4,4] is good.

• The subarray consists of exactly 3 consecutive increasing elements, that is, the difference between adjacent elements is 1. For example, the subarray [3,4,5] is good, but the subarray [1,3,5] is not.

Return true** if the array has at least one valid partition**. Otherwise, return false.

  Example 1:

Input: nums = [4,4,4,5,6]
Output: true
Explanation: The array can be partitioned into the subarrays [4,4] and [4,5,6].
This partition is valid, so we return true.

Example 2:

Input: nums = [1,1,1,2]
Output: false
Explanation: There is no valid partition for this array.

  Constraints:

• 2 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^6

Solution (Java)

class Solution {
    public boolean validPartition(int[] nums) {
        boolean[] canPartition = new boolean[nums.length + 1];
        canPartition[0] = true;
        int diff = nums[1] - nums[0];
        boolean equal = diff == 0;
        boolean incOne = diff == 1;
        canPartition[2] = equal;
        for (int i = 3; i < canPartition.length; i++) {
            diff = nums[i - 1] - nums[i - 2];
            if (diff == 0) {
                canPartition[i] = canPartition[i - 2] || (equal && canPartition[i - 3]);
                equal = true;
                incOne = false;
            } else if (diff == 1) {
                canPartition[i] = incOne && canPartition[i - 3];
                equal = false;
                incOne = true;
            }
        }
        return canPartition[nums.length];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).