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

Python Code
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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)

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

Go Code
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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)
}

TypeScript Code
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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;
}

Solution 2

Python Code
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class Solution:
    def validPartition(self, nums: List[int]) -> bool:
        n = len(nums)
        dp = [False] * (n + 1)
        dp[0] = True
        for i in range(2, n + 1):
            if nums[i - 1] == nums[i - 2]:
                dp[i] = dp[i] or dp[i - 2]
            if i > 2 and nums[i - 1] == nums[i - 2] == nums[i - 3]:
                dp[i] = dp[i] or dp[i - 3]
            if (
                i > 2
                and nums[i - 1] - nums[i - 2] == 1
                and nums[i - 2] - nums[i - 3] == 1
            ):
                dp[i] = dp[i] or dp[i - 3]
        return dp[-1]

Java Code
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class Solution {
    public boolean validPartition(int[] nums) {
        int n = nums.length;
        boolean[] dp = new boolean[n + 1];
        dp[0] = true;
        for (int i = 2; i <= n; ++i) {
            if (nums[i - 1] == nums[i - 2]) {
                dp[i] = dp[i] || dp[i - 2];
            }
            if (i > 2 && nums[i - 1] == nums[i - 2] && nums[i - 2] == nums[i - 3]) {
                dp[i] = dp[i] || dp[i - 3];
            }
            if (i > 2 && nums[i - 1] - nums[i - 2] == 1 && nums[i - 2] - nums[i - 3] == 1) {
                dp[i] = dp[i] || dp[i - 3];
            }
        }
        return dp[n];
    }
}

C++ Code
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class Solution {
public:
    bool validPartition(vector<int>& nums) {
        int n = nums.size();
        vector<bool> dp(n + 1);
        dp[0] = true;
        for (int i = 2; i <= n; ++i) {
            if (nums[i - 1] == nums[i - 2]) dp[i] = dp[i] || dp[i - 2];
            if (i > 2 && nums[i - 1] == nums[i - 2] && nums[i - 2] == nums[i - 3]) dp[i] = dp[i] || dp[i - 3];
            if (i > 2 && nums[i - 1] - nums[i - 2] == 1 && nums[i - 2] - nums[i - 3] == 1) dp[i] = dp[i] || dp[i - 3];
        }
        return dp[n];
    }
};

Go Code
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func validPartition(nums []int) bool {
	n := len(nums)
	dp := make([]bool, n+1)
	dp[0] = true
	for i := 2; i <= n; i++ {
		if nums[i-1] == nums[i-2] {
			dp[i] = dp[i] || dp[i-2]
		}
		if i > 2 && nums[i-1] == nums[i-2] && nums[i-2] == nums[i-3] {
			dp[i] = dp[i] || dp[i-3]
		}
		if i > 2 && nums[i-1]-nums[i-2] == 1 && nums[i-2]-nums[i-3] == 1 {
			dp[i] = dp[i] || dp[i-3]
		}
	}
	return dp[n]
}

TypeScript Code
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function validPartition(nums: number[]): boolean {
    const n = nums.length;
    const dp = new Array(n + 1).fill(false);
    dp[0] = true;
    for (let i = 2; i <= n; ++i) {
        if (nums[i - 1] == nums[i - 2]) {
            dp[i] = dp[i] || dp[i - 2];
        }
        if (i > 2 && nums[i - 1] == nums[i - 2] && nums[i - 2] == nums[i - 3]) {
            dp[i] = dp[i] || dp[i - 3];
        }
        if (i > 2 && nums[i - 1] - nums[i - 2] == 1 && nums[i - 2] - nums[i - 3] == 1) {
            dp[i] = dp[i] || dp[i - 3];
        }
    }
    return dp[n];
}