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

Description

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.

 

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

Solutions

Solution 1: Memoization Search

We design a function $dfs(i)$, which represents whether there is a valid partition starting from index $i$. So the answer is $dfs(0)$.

The execution process of the function $dfs(i)$ is as follows:

• If $i \ge n$, return $true$.
• If the elements at index $i$ and $i+1$ are equal, we can choose to make $i$ and $i+1$ a subarray, and recursively call $dfs(i+2)$.
• If the elements at index $i$, $i+1$ and $i+2$ are equal, we can choose to make $i$, $i+1$ and $i+2$ a subarray, and recursively call $dfs(i+3)$.
• If the elements at index $i$, $i+1$ and $i+2$ increase by $1$ in turn, we can choose to make $i$, $i+1$ and $i+2$ a subarray, and recursively call $dfs(i+3)$.
• If none of the above conditions are met, return $false$, otherwise return $true$.

That is:

\[dfs(i) = \textit{OR} \begin{cases} true,&i \ge n\\ dfs(i+2),&i+1 < n\ \textit{and}\ \textit{nums}[i] = \textit{nums}[i+1]\\ dfs(i+3),&i+2 < n\ \textit{and}\ \textit{nums}[i] = \textit{nums}[i+1] = \textit{nums}[i+2]\\ dfs(i+3),&i+2 < n\ \textit{and}\ \textit{nums}[i+1] - \textit{nums}[i] = 1\ \textit{and}\ \textit{nums}[i+2] - \textit{nums}[i+1] = 1 \end{cases}\]

To avoid repeated calculations, we use the method of memoization search.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming.

Let $f[i]$ represent whether there is a valid partition for the first $i$ elements of the array. Initially, $f[0] = true$, and the answer is $f[n]$.

The state transition equation is as follows:

\[f[i] = \textit{OR} \begin{cases} true,&i = 0\\ f[i-2],&i-2 \ge 0\ \textit{and}\ \textit{nums}[i-1] = \textit{nums}[i-2]\\ f[i-3],&i-3 \ge 0\ \textit{and}\ \textit{nums}[i-1] = \textit{nums}[i-2] = \textit{nums}[i-3]\\ f[i-3],&i-3 \ge 0\ \textit{and}\ \textit{nums}[i-1] - \textit{nums}[i-2] = 1\ \textit{and}\ \textit{nums}[i-2] - \textit{nums}[i-3] = 1 \end{cases}\]

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      private int n;
      private int[] f;
      private int[] nums;
  
      public boolean validPartition(int[] nums) {
          this.nums = nums;
          n = nums.length;
          f = new int[n];
          Arrays.fill(f, -1);
          return dfs(0);
      }
  
      private boolean dfs(int i) {
          if (i == n) {
              return true;
          }
          if (f[i] != -1) {
              return f[i] == 1;
          }
          boolean res = false;
          if (i < n - 1 && nums[i] == nums[i + 1]) {
              res = res || dfs(i + 2);
          }
          if (i < n - 2 && nums[i] == nums[i + 1] && nums[i + 1] == nums[i + 2]) {
              res = res || dfs(i + 3);
          }
          if (i < n - 2 && nums[i + 1] - nums[i] == 1 && nums[i + 2] - nums[i + 1] == 1) {
              res = res || dfs(i + 3);
          }
          f[i] = res ? 1 : 0;
          return res;
      }
  }
  
  
  // Solution 2
  class Solution {
      public boolean validPartition(int[] nums) {
          int n = nums.length;
          boolean[] f = new boolean[n + 1];
          f[0] = true;
          for (int i = 1; i <= n; ++i) {
              boolean a = i - 2 >= 0 && nums[i - 1] == nums[i - 2];
              boolean b = i - 3 >= 0 && nums[i - 1] == nums[i - 2] && nums[i - 2] == nums[i - 3];
              boolean c
                  = i - 3 >= 0 && nums[i - 1] - nums[i - 2] == 1 && nums[i - 2] - nums[i - 3] == 1;
              f[i] = (a && f[i - 2]) || ((b || c) && f[i - 3]);
          }
          return f[n];
      }
  }
  
  

• class Solution {
  public:
      vector<int> f;
      vector<int> nums;
      int n;
  
      bool validPartition(vector<int>& nums) {
          n = nums.size();
          this->nums = nums;
          f.assign(n, -1);
          return dfs(0);
      }
  
      bool dfs(int i) {
          if (i == n) return true;
          if (f[i] != -1) return f[i] == 1;
          bool res = false;
          if (i < n - 1 && nums[i] == nums[i + 1]) res = res || dfs(i + 2);
          if (i < n - 2 && nums[i] == nums[i + 1] && nums[i + 1] == nums[i + 2]) res = res || dfs(i + 3);
          if (i < n - 2 && nums[i + 1] - nums[i] == 1 && nums[i + 2] - nums[i + 1] == 1) res = res || dfs(i + 3);
          f[i] = res ? 1 : 0;
          return res;
      }
  };
  
  
  // Solution 2
  class Solution {
  public:
      bool validPartition(vector<int>& nums) {
          int n = nums.size();
          vector<bool> f(n + 1);
          f[0] = true;
          for (int i = 1; i <= n; ++i) {
              bool a = i - 2 >= 0 && nums[i - 1] == nums[i - 2];
              bool b = i - 3 >= 0 && nums[i - 1] == nums[i - 2] && nums[i - 2] == nums[i - 3];
              bool c = i - 3 >= 0 && nums[i - 1] - nums[i - 2] == 1 && nums[i - 2] - nums[i - 3] == 1;
              f[i] = (a && f[i - 2]) || ((b || c) && f[i - 3]);
          }
          return f[n];
      }
  };
  
  

• class Solution:
      def validPartition(self, nums: List[int]) -> bool:
          @cache
          def dfs(i):
              if i == n:
                  return True
              res = False
              if i < n - 1 and nums[i] == nums[i + 1]:
                  res = res or dfs(i + 2)
              if i < n - 2 and nums[i] == nums[i + 1] and nums[i + 1] == nums[i + 2]:
                  res = res or dfs(i + 3)
              if (
                  i < n - 2
                  and nums[i + 1] - nums[i] == 1
                  and nums[i + 2] - nums[i + 1] == 1
              ):
                  res = res or dfs(i + 3)
              return res
  
          n = len(nums)
          return dfs(0)
  
  
  # Solution 2
  class Solution:
      def validPartition(self, nums: List[int]) -> bool:
          n = len(nums)
          f = [True] + [False] * n
          for i, x in enumerate(nums, 1):
              a = i - 2 >= 0 and nums[i - 2] == x
              b = i - 3 >= 0 and nums[i - 3] == nums[i - 2] == x
              c = i - 3 >= 0 and x - nums[i - 2] == 1 and nums[i - 2] - nums[i - 3] == 1
              f[i] = (a and f[i - 2]) or ((b or c) and f[i - 3])
          return f[n]
  
  

• func validPartition(nums []int) bool {
  	n := len(nums)
  	f := make([]int, n)
  	for i := range f {
  		f[i] = -1
  	}
  	var dfs func(int) bool
  	dfs = func(i int) bool {
  		if i == n {
  			return true
  		}
  		if f[i] != -1 {
  			return f[i] == 1
  		}
  		res := false
  		f[i] = 0
  		if i < n-1 && nums[i] == nums[i+1] {
  			res = res || dfs(i+2)
  		}
  		if i < n-2 && nums[i] == nums[i+1] && nums[i+1] == nums[i+2] {
  			res = res || dfs(i+3)
  		}
  		if i < n-2 && nums[i+1]-nums[i] == 1 && nums[i+2]-nums[i+1] == 1 {
  			res = res || dfs(i+3)
  		}
  		if res {
  			f[i] = 1
  		}
  		return res
  	}
  	return dfs(0)
  }
  
  
  // Solution 2
  func validPartition(nums []int) bool {
  	n := len(nums)
  	f := make([]bool, n+1)
  	f[0] = true
  	for i := 1; i <= n; i++ {
  		x := nums[i-1]
  		a := i-2 >= 0 && nums[i-2] == x
  		b := i-3 >= 0 && nums[i-3] == nums[i-2] && nums[i-2] == x
  		c := i-3 >= 0 && x-nums[i-2] == 1 && nums[i-2]-nums[i-3] == 1
  		f[i] = (a && f[i-2]) || ((b || c) && f[i-3])
  	}
  	return f[n]
  }
  
  

• function validPartition(nums: number[]): boolean {
      const n = nums.length;
      const vis = new Array(n).fill(false);
      const queue = [0];
      while (queue.length !== 0) {
          const i = queue.shift() ?? 0;
  
          if (i === n) {
              return true;
          }
  
          if (!vis[i + 2] && i + 2 <= n && nums[i] === nums[i + 1]) {
              queue.push(i + 2);
              vis[i + 2] = true;
          }
  
          if (
              !vis[i + 3] &&
              i + 3 <= n &&
              ((nums[i] === nums[i + 1] && nums[i + 1] === nums[i + 2]) ||
                  (nums[i] === nums[i + 1] - 1 && nums[i + 1] === nums[i + 2] - 1))
          ) {
              queue.push(i + 3);
              vis[i + 3] = true;
          }
      }
      return false;
  }
  
  
  // Solution 2
  function validPartition(nums: number[]): boolean {
      const n = nums.length;
      const f: boolean[] = Array(n + 1).fill(false);
      f[0] = true;
      for (let i = 1; i <= n; ++i) {
          const a = i - 2 >= 0 && nums[i - 1] === nums[i - 2];
          const b = i - 3 >= 0 && nums[i - 1] === nums[i - 2] && nums[i - 2] === nums[i - 3];
          const c = i - 3 >= 0 && nums[i - 1] - nums[i - 2] === 1 && nums[i - 2] - nums[i - 3] === 1;
          f[i] = (a && f[i - 2]) || ((b || c) && f[i - 3]);
      }
      return f[n];
  }
  
  

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