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Diffstat (limited to 'other/check_valid_partition.cpp')
-rw-r--r-- | other/check_valid_partition.cpp | 50 |
1 files changed, 50 insertions, 0 deletions
diff --git a/other/check_valid_partition.cpp b/other/check_valid_partition.cpp new file mode 100644 index 0000000..3f4cd76 --- /dev/null +++ b/other/check_valid_partition.cpp @@ -0,0 +1,50 @@ +// 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 +} |