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// 2369. Check if There is a Valid Partition For The Array
#include "leetcode.h"
/*
* 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.
*/
class Solution {
bool solve(vector<int> &nums, int i, vector<int> &dp) {
if (nums.size() == i)
return true;
if (dp[i] != -1)
return dp[i];
if (i + 1 < nums.size() && nums[i] == nums[i + 1]) {
if (solve(nums, i + 2, dp))
return true;
if (i + 2 < nums.size() && nums[i] == nums[i + 2])
if (solve(nums, i + 3, dp))
return true;
}
if (i + 2 < nums.size() && nums[i] == nums[i + 1] - 1 &&
nums[i] == nums[i + 2] - 2)
if (solve(nums, i + 3, dp))
return true;
return !dp[i];
}
public:
bool validPartition(vector<int> &nums) {
vector<int> dp(nums.size(), -1);
return solve(nums, 0, dp);
}
};
int main() {
Solution obj;
vector<int> n1 = {4, 4, 4, 5, 6}, n2 = {1, 1, 1, 2};
printf("%d\n", obj.validPartition(n1)); // expect: 1
printf("%d\n", obj.validPartition(n2)); // expect: 0
}
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