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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
+}