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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:
- 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 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 <= 105
1 <= nums[i] <= 106
Solutions
-
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; } }
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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; } };
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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)
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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) }
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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; }