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 <= 105

• 1 <= nums[i] <= 106

Solution

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var validPartition = function(nums) {
    var dp = Array(nums.length + 1);
    dp[nums.length] = true;
    for (var i = nums.length - 1; i >= 0; i--) {
        dp[i] = (nums[i] === nums[i + 1] && dp[i + 2])
            || (nums[i] === nums[i + 1] && nums[i + 1] === nums[i + 2] && dp[i + 3])
            || (nums[i] + 1 === nums[i + 1] && nums[i + 1] + 1 === nums[i + 2] && dp[i + 3]);
    }
    return dp[0];
};

Explain:

nope.

Complexity:

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