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:
- 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
3consecutive increasing elements, that is, the difference between adjacent elements is1. 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 <= 1051 <= 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.
Python3
class Solution:
def validPartition(self, nums: List[int]) -> bool:
@cache
def dfs(i: int) -> bool:
if i >= n:
return True
a = i + 1 < n and nums[i] == nums[i + 1]
b = i + 2 < n and nums[i] == nums[i + 1] == nums[i + 2]
c = (
i + 2 < n
and nums[i + 1] - nums[i] == 1
and nums[i + 2] - nums[i + 1] == 1
)
return (a and dfs(i + 2)) or ((b or c) and dfs(i + 3))
n = len(nums)
return dfs(0)
Java
class Solution {
private int n;
private int[] nums;
private Boolean[] f;
public boolean validPartition(int[] nums) {
n = nums.length;
this.nums = nums;
f = new Boolean[n];
return dfs(0);
}
private boolean dfs(int i) {
if (i >= n) {
return true;
}
if (f[i] != null) {
return f[i];
}
boolean a = i + 1 < n && nums[i] == nums[i + 1];
boolean b = i + 2 < n && nums[i] == nums[i + 1] && nums[i + 1] == nums[i + 2];
boolean c = i + 2 < n && nums[i + 1] - nums[i] == 1 && nums[i + 2] - nums[i + 1] == 1;
return f[i] = ((a && dfs(i + 2)) || ((b || c) && dfs(i + 3)));
}
}
C++
class Solution {
public:
bool validPartition(vector<int>& nums) {
n = nums.size();
this->nums = nums;
f.assign(n, -1);
return dfs(0);
}
private:
int n;
vector<int> f;
vector<int> nums;
bool dfs(int i) {
if (i >= n) {
return true;
}
if (f[i] != -1) {
return f[i] == 1;
}
bool a = i + 1 < n && nums[i] == nums[i + 1];
bool b = i + 2 < n && nums[i] == nums[i + 1] && nums[i + 1] == nums[i + 2];
bool c = i + 2 < n && nums[i + 1] - nums[i] == 1 && nums[i + 2] - nums[i + 1] == 1;
f[i] = ((a && dfs(i + 2)) || ((b || c) && dfs(i + 3))) ? 1 : 0;
return f[i] == 1;
}
};
Go
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
}
a := i+1 < n && nums[i] == nums[i+1]
b := i+2 < n && nums[i] == nums[i+1] && nums[i+1] == nums[i+2]
c := i+2 < n && nums[i+1]-nums[i] == 1 && nums[i+2]-nums[i+1] == 1
f[i] = 0
if a && dfs(i+2) || b && dfs(i+3) || c && dfs(i+3) {
f[i] = 1
}
return f[i] == 1
}
return dfs(0)
}
TypeScript
function validPartition(nums: number[]): boolean {
const n = nums.length;
const f: number[] = Array(n).fill(-1);
const dfs = (i: number): boolean => {
if (i >= n) {
return true;
}
if (f[i] !== -1) {
return f[i] === 1;
}
const a = i + 1 < n && nums[i] == nums[i + 1];
const b = i + 2 < n && nums[i] == nums[i + 1] && nums[i + 1] == nums[i + 2];
const c = i + 2 < n && nums[i + 1] - nums[i] == 1 && nums[i + 2] - nums[i + 1] == 1;
f[i] = (a && dfs(i + 2)) || ((b || c) && dfs(i + 3)) ? 1 : 0;
return f[i] == 1;
};
return dfs(0);
}
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.
Python3
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]
Java
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];
}
}
C++
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];
}
};
Go
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]
}
TypeScript
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];
}