The trick here is to name the state correctly, then let the implementation follow. For 1569. Number of Ways to Reorder Array to Get Same BST, the solution in this repository is mainly a two pointers solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: two pointers, sliding window.
The notes already sitting in the source point us in the right direction:
- https://stackoverflow.com/questions/17119116/how-many-ways-can-you-insert-a-series-of-values-into-a-bst-to-form-a-specific-tr
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are insert, dfs, fact, power, gcd.
Guide
Why?
The solution works because it narrows the problem until every update has a clear reason to exist.
- A map keeps the lookup side cheap; the code pays a little memory to avoid repeated searching.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//https://stackoverflow.com/questions/17119116/how-many-ways-can-you-insert-a-series-of-values-into-a-bst-to-form-a-specific-tr
02//Runtime: 128 ms, faster than 80.00% of C++ online submissions for Number of Ways to Reorder Array to Get Same BST.
03//Memory Usage: 28.5 MB, less than 100.00% of C++ online submissions for Number of Ways to Reorder Array to Get Same BST.
04class Node{
05public:
06 Node *left, *right;
07 int val;
08
09 Node(int v) : val(v), left(nullptr), right(nullptr){};
10};
11
12class BST{
13public:
14 Node* root;
15
16 void insert(int v, Node* node = nullptr){
17 if(root == nullptr){
18 root = new Node(v);
19 }else{
20 if(node == nullptr)
21 node = root;
22 if(v < node->val){
23 if(node->left == nullptr){
24 // cout << node->val << " l-> " << v << endl;
25 node->left = new Node(v);
26 }else{
27 // cout << node->left->val << " l-> " << v << endl;
28 insert(v, node->left);
29 }
30 }else{
31 if(node->right == nullptr){
32 // cout << node->val << " r-> " << v << endl;
33 node->right = new Node(v);
34 }else{
35 // cout << node->right->val << " r-> " << v << endl;
36 insert(v, node->right);
37 }
38 }
39 }
40 }
41};
42
43
44class Solution {
45public:
46 int MOD = 1e9+7;
47 unordered_map<Node*, int> counter;
48
49 //fill "counter"
50 int dfs(Node* node){
51 if(node == nullptr) return 0;
52 counter[node] = dfs(node->left) + dfs(node->right) + 1;
53 return counter[node];
54 };
55
56 // Returns factorial of n
57 int fact(int n)
58 {
59 long long res = 1;
60 for (int i = 2; i <= n; i++)
61 res = (res * i) % MOD;
62 return res;
63 };
64
65 int power(int x, unsigned int y){
66 if (y == 0)
67 return 1;
68 long long p = power(x, y/2) % MOD;
69 p = (p * p) % MOD;
70
71 return (y%2 == 0)? p : (x * p) % MOD;
72 };
73
74 // Function to return gcd of a and b
75 int gcd(int a, int b){
76 if (a == 0)
77 return b;
78 return gcd(b%a, a);
79 };
80
81 // Function to find modular inverse of a under modulo m
82 // Assumption: m is prime
83 int modInverse(int a){
84 int g = gcd(a, MOD);
85 if (g != 1){
86 //Fermat's little theorem only works only when m is prime!
87 return -1;
88 }else{
89 // If a and m are relatively prime, then modulo inverse
90 // is a^(m-2) mod m
91 /*
92 From Fermat's little theorem,
93 when a is not divisible by p,
94 a^(p-1) mod m = 1
95 here we choose p as m,
96 so a^(m-1) mod m = 1
97 this is equal to saying that
98 a * a^(m-2) mod m = 1
99 so we can say that a^(m-2) mod m is the inverse of a
100 */
101 return power(a, MOD-2);
102 }
103 };
104
105 int nCr(int n, int r)
106 {
107 long long fr_inverse = modInverse(fact(r));
108 long long fnr_inverse = modInverse(fact(n - r));
109 return ((fact(n) * fr_inverse) % MOD * fnr_inverse) % MOD;
110 };
111
112 int countInsertionOrderings(Node* node) {
113 if(!node) return 1;
114
115 int m = counter[node->left];
116 int n = counter[node->right];
117
118 //number of ways to permute sequence of L and sequence of R
119 long long a = nCr(m + n, n);
120 //how many possible permutations of sequence L
121 long long b = countInsertionOrderings(node->left);
122 //how many possible permutations of sequence R
123 long long c = countInsertionOrderings(node->right);
124
125 return (((a*b) % MOD) * c) % MOD;
126 }
127
128 int numOfWays(vector<int>& nums) {
129 int n = nums.size();
130 if(n == 0) return 0;
131
132 BST* bst = new BST();
133
134 for(int i = 0; i < n; ++i){
135 bst->insert(nums[i]);
136 }
137
138 dfs(bst->root);
139
140 //-1: deduce one, the original way to construct binary tree
141 return countInsertionOrderings(bst->root) - 1;
142 }
143};
144
145//use pascal triangle to calculate combination count
146//https://leetcode.com/problems/number-of-ways-to-reorder-array-to-get-same-bst/discuss/819369/C%2B%2B-Just-using-recursion-very-Clean-and-Easy-to-understand-O(n2)
147//https://www.geeksforgeeks.org/calculate-ncr-using-pascals-triangle/
148//Runtime: 376 ms, faster than 37.46% of C++ online submissions for Number of Ways to Reorder Array to Get Same BST.
149//Memory Usage: 105.7 MB, less than 38.77% of C++ online submissions for Number of Ways to Reorder Array to Get Same BST.
150class Solution {
151public:
152 long long MOD = 1e9+7;
153 vector<vector<int>> pascal;
154
155 void build_pascal(int max_num){
156 pascal = vector<vector<int>>(max_num+1);
157
158 for(int n = 0; n <= max_num; ++n){
159 //we need nC0 to nCn
160 pascal[n] = vector<int>(n+1);
161 //cout << "pascal[" << n << "]" << endl;
162
163 /*
164 to calculate pascal[n][n],
165 it will use pascal[n-1][n], which doesn't exist,
166 so here we treat pascal[n][n] as a special case
167 and do not calculate it in the loop
168 */
169 pascal[n][0] = pascal[n][n] = 1;
170 for(int r = 1; r < n; ++r){
171 pascal[n][r] = (pascal[n-1][r-1] + pascal[n-1][r]) % MOD;
172 }
173 }
174 }
175
176 int dfs(vector<int>& nums){
177 int n = nums.size();
178 if(n == 0) return 1;
179
180 vector<int> left, right;
181
182 //i starts from 1: exclude nums[0], which is root
183 for(int i = 1; i < n; ++i){
184 if(nums[i] < nums[0]) left.push_back(nums[i]);
185 else right.push_back(nums[i]);
186 }
187
188 long long left_res = dfs(left);
189 long long right_res = dfs(right);
190
191 //(n-1)C(left.size()): choose from n-1 elements, excluding the root
192 return ((pascal[n-1][left.size()] * left_res) % MOD * right_res) % MOD;
193 }
194
195 int numOfWays(vector<int>& nums) {
196 int n = nums.size();
197 build_pascal(n);
198 //cout << "table built" << endl;
199
200 return dfs(nums)-1;
201 }
202};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.