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Formatted question description: https://leetcode.ca/all/2369.html

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

• Difficulty: Medium.
• Related Topics: Array, Dynamic Programming.
• Similar Questions: .

## 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 (Java, C++, Python)

• class Solution {
public boolean validPartition(int[] nums) {
boolean[] canPartition = new boolean[nums.length + 1];
canPartition[0] = true;
int diff = nums[1] - nums[0];
boolean equal = diff == 0;
boolean incOne = diff == 1;
canPartition[2] = equal;
for (int i = 3; i < canPartition.length; i++) {
diff = nums[i - 1] - nums[i - 2];
if (diff == 0) {
canPartition[i] = canPartition[i - 2] || (equal && canPartition[i - 3]);
equal = true;
incOne = false;
} else if (diff == 1) {
canPartition[i] = incOne && canPartition[i - 3];
equal = false;
incOne = true;
}
}
return canPartition[nums.length];
}
}

############

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;
}
}

• 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)

############

# 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/

class Solution:
def validPartition(self, nums: List[int]) -> bool:
n = len(nums)

@cache
def go(index):
if index >= n:
return True

valid = False

if index + 1 < n and nums[index] == nums[index + 1]:
valid |= go(index + 2)

if index + 2 < n and nums[index] == nums[index + 1] == nums[index + 2]:
valid |= go(index + 3)

if index + 2 < n and nums[index] + 1 == nums[index + 1] and nums[index + 1] + 1 == nums[index + 2]:
valid |= go(index + 3)

return valid

return go(0)


• 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;
}
};

• 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)
}

• 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;
}



Explain:

nope.

Complexity:

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