2369

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

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

Companies: Google

Related Topics:
Array, Dynamic Programming

Solution 1. Top-down DP

// OJ: https://leetcode.com/problems/check-if-there-is-a-valid-partition-for-the-array
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool validPartition(vector<int>& A) {
        int N = A.size();
        vector<int> memo(N, -1);
        function<bool(int)> dfs = [&](int start) -> bool {
            if (start == N) return true;
            if (memo[start] != -1) return memo[start];
            bool ans = false;
            if (start + 1 < N && A[start + 1] == A[start]) ans = dfs(start + 2);
            if (start + 2 < N &&
                ((A[start + 1] == A[start] && A[start + 2] == A[start])
                || (A[start + 1] == A[start] + 1 && A[start + 2] == A[start] + 2))) ans |= dfs(start + 3);
            return memo[start] = ans;
        };
        return dfs(0);
    }
};