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

Problem Statement

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

Intuition

Just input the three given conditions in recursion and memoize

Links

https://leetcode.com/problems/check-if-there-is-a-valid-partition-for-the-array/description/

Video Links

Approach 1:

Memoization

C++

class Solution {
public:
    bool find_ans(vector<int>& nums, int index, vector<int> &dp){
        if(index == nums.size())
            return true;

        if(dp[index] != -1)
            return dp[index];

        if(index+1< nums.size() and nums[index] == nums[index+1]){
            if( find_ans(nums, index+2, dp) )
                return dp[index] = true;

            if(index+2< nums.size() and nums[index] == nums[index+2])
                if( find_ans(nums, index+3, dp) )
                    return dp[index] = true;
        }

        if(index+2 < nums.size() and nums[index] == nums[index+1]-1 and nums[index] == nums[index+2]-2){
            if( find_ans(nums, index+3, dp) )
                return dp[index] = true;
        }

        return dp[index] = false;
    }

    bool validPartition(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n, -1);

        return find_ans(nums, 0, dp);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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