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).

2369 - Check if There is a Valid Partition For The Array

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

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Solution (Java, C++, Python)

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

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Solution (Java, C++, Python)

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