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-rw-r--r--check_valid_partition.cpp50
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diff --git a/check_valid_partition.cpp b/check_valid_partition.cpp
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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
-}