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-rw-r--r--other/num_ways_reorder_array.cpp57
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diff --git a/other/num_ways_reorder_array.cpp b/other/num_ways_reorder_array.cpp
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-// 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
-}