Check if There is a Valid Partition For The Array - Two Approaches - DP & Recursion w/ Memoization [JS]

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Description 

Solution 1: DP - Recursion w/ Memoization

Memoize each dp(i), where dp(i) = whether it is possible to have a valid partition from index i to n.
For each index i, try to make it part of a valid partition (one of the three cases):

  1. Two equal elements
  2. Three equal elements
  3. Three consecutive elements

Time Complexity: O(n) 158ms
Space Complexity: O(n) 59.9MB

var validPartition = function(nums) {
  let n = nums.length, memo = Array(n).fill(-1);
  return dp(0);

  function dp(i) {
    if (i === n) return true;
    if (i === n - 1) return false;
    if (memo[i] !== -1) return memo[i];

    if (nums[i] === nums[i + 1] && dp(i + 2)) return memo[i] = true;
    if (i < n - 2) {
      if (!dp(i + 3)) return memo[i] = false;
      let hasThreeEqual = nums[i] === nums[i + 1] && nums[i + 1] === nums[i + 2];
      let hasThreeConsecutive = nums[i] + 1 === nums[i + 1] && nums[i + 1] + 1 === nums[i + 2];
      if (hasThreeEqual || hasThreeConsecutive) return memo[i] = true;
    }
    return memo[i] = false;
  }  
};

Solution 2: DP - Tabulation

We can save a bit of space by using iteration instead of recursion.

Time Complexity: O(n) 122ms
Space Complexity: O(n) 54.3MB

var validPartition = function(nums) {
  let n = nums.length, dp = Array(n + 1).fill(false);
  dp[n] = true;
  for (let i = n - 2; i >= 0; i--) {
    if (nums[i] === nums[i + 1] && dp[i + 2]) dp[i] = true;
    else if (i < n - 2) {
      if (!dp[i + 3]) continue;
      let hasThreeEqual = nums[i] === nums[i + 1] && nums[i + 1] === nums[i + 2];
      let hasThreeConsecutive = nums[i] + 1 === nums[i + 1] && nums[i + 1] + 1 === nums[i + 2];
      if (hasThreeEqual || hasThreeConsecutive) dp[i] = true;
    }
  }
  return dp[0];
};

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