2369-check-if-there-is-a-valid-partition-for-the-array.js

/**
 * 2369. Check if There is a Valid Partition For The Array
 * https://leetcode.com/problems/check-if-there-is-a-valid-partition-for-the-array/
 * Difficulty: Medium
 *
 * 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:
 * 1. The subarray consists of exactly 2, equal elements. For example, the subarray [2,2] is good.
 * 2. The subarray consists of exactly 3, equal elements. For example, the subarray [4,4,4] is good.
 * 3. 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.
 */

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var validPartition = function(nums) {
  const n = nums.length;
  const dp = new Array(n + 1).fill(false);
  dp[0] = true;

  for (let i = 2; i <= n; i++) {
    if (nums[i - 1] === nums[i - 2]) {
      dp[i] = dp[i] || dp[i - 2];
    }
    if (i >= 3) {
      if (nums[i - 1] === nums[i - 2] && nums[i - 2] === nums[i - 3]) {
        dp[i] = dp[i] || dp[i - 3];
      }
      if (nums[i - 1] === nums[i - 2] + 1 && nums[i - 2] === nums[i - 3] + 1) {
        dp[i] = dp[i] || dp[i - 3];
      }
    }
  }

  return dp[n];
};