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Diffstat (limited to 'other/num_ways_reorder_array.cpp')
-rw-r--r-- | other/num_ways_reorder_array.cpp | 57 |
1 files changed, 57 insertions, 0 deletions
diff --git a/other/num_ways_reorder_array.cpp b/other/num_ways_reorder_array.cpp new file mode 100644 index 0000000..9adcb55 --- /dev/null +++ b/other/num_ways_reorder_array.cpp @@ -0,0 +1,57 @@ +// 1569. Number of Ways to Reorder Array to Get Same BST +#include "leetcode.h" + +/* + * given an array 'nums' that represents a permutation of integers from '1 to + * n'. we are going to construct a binary search tree (BST) by inserting the + * elements of 'nums' in order into an initially empty BST. find the number of + * different ways to reorder 'nums' so that the constructed BST is identical to + * that formed from the original array 'nums'. for example given 'nums = + * [2,1,3]', we will have 2 as the root, 1 as a left child, and three as a right + * child. the array [2,3,1] also yields the same BST, but '[3,2,1]' yields a + * different BST. return the number of ways to reorder 'nums' such that the BST + * formed is identical to the original BST formed from 'nums'. since the answer + * may be large, return it modulo 10^9+7 + */ + +class Solution { + int mod = 1e+9 + 7; + long l[10001][10001] = {0}; + bool init = false; + long combination(int n, int r) { + if (init) + return l[n][r]; + l[0][0] = 1; + for (int i = 1; i < 1001; i++) { + l[i][0] = 1; + for (int j = 1; j < i + 1; j++) + l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]) % mod; + } + init = true; + return l[n][r]; + } + long dfs(vector<int> &nums) { + if (nums.size() < 3) + return 1; + vector<int> l, r; + for (auto n : nums) { + if (n > nums[0]) + l.push_back(n); + if (n < nums[0]) + r.push_back(n); + } + long c = combination(l.size() + r.size(), l.size()); + return c * dfs(l) % mod * dfs(r) % mod; + } + +public: + int numOfWays(vector<int> &nums) { return dfs(nums) - 1; } +}; + +int main() { + Solution obj; + vector<int> n1 = {2, 1, 3}, n2 = {3, 4, 5, 1, 2}, n3 = {1, 2, 3}; + printf("%d\n", obj.numOfWays(n1)); // expect: 1 + printf("%d\n", obj.numOfWays(n2)); // expect: 5 + printf("%d\n", obj.numOfWays(n3)); // expect: 0 +} |