A good way into this one is to ask: what do we need to remember so we never redo work blindly? For 730. Count Different Palindromic Subsequences, the solution in this repository is mainly a dynamic programming solution.
Guide
What?
The first job is to translate the English prompt into state, transition, and stopping conditions. 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: dynamic programming, two pointers, sliding window.
The notes already sitting in the source point us in the right direction:
- DP, O(N^3)
- https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109507/Java-96ms-DP-Solution-with-Detailed-Explanation
- time: O(N^3), space: O(N^2)
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 countPalindromicSubsequences, nextRight, char2pos.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- 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^3), space: O(N^2)
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//DP, O(N^3)
02//https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109507/Java-96ms-DP-Solution-with-Detailed-Explanation
03//Runtime: 164 ms, faster than 60.90% of C++ online submissions for Count Different Palindromic Subsequences.
04//Memory Usage: 36.4 MB, less than 100.00% of C++ online submissions for Count Different Palindromic Subsequences.
05//time: O(N^3), space: O(N^2)
06class Solution {
07public:
08 int countPalindromicSubsequences(string S) {
09 int n = S.size();
10 const int MOD = 1e9+7;
11 // vector<vector<long long>> dp(n, vector(n, 0LL)); //don't need this!!
12 vector<vector<int>> dp(n, vector(n, 0));
13
14 for(int dist = 0; dist < n; dist++){
15 for(int i = 0; i + dist < n; i++){
16 int j = i+dist;
17 //dp[i][j]: count of palindrome in S[i...j]
18 if(i == j){
19 //base case: only one char
20 dp[i][j] = 1;
21 continue;
22 }
23
24 if(S[i] != S[j]){
25 /*
26 dp[i+1][j-1] is included in both
27 dp[i][j-1] and dp[i+1][j],
28 so we need to deduce it
29 */
30 dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];
31 }else{
32 //S[i] == S[j]
33 int lo = i+1, hi = j-1;
34
35 /*
36 try to find S[i] in S[i+1...j-1],
37 if found,
38 S[lo] will be the first S[i] in S[i+1...j-1]
39 if not found,
40 lo will > hi
41 */
42 while(lo <= hi && S[lo] != S[i]){
43 lo++;
44 }
45
46 /*
47 either found: S[hi] will be S[i]
48 or not found: hi < lo
49 */
50 while(lo <= hi && S[hi] != S[i]){
51 hi--;
52 }
53
54 /*
55 now there are 3 cases:
56 lo > hi: not found S[i] in S[i+1...j-1]
57 lo == hi: only one S[i] in S[i+1...j-1]
58 lo < hi: at least two S[i] in S[i+1...j-1]
59 */
60
61 if(lo > hi){
62 /*
63 axxxa
64 dp[i+1][j-1]*2:
65 both the palindrome count in xxx itself and
66 axxxa, so two times
67
68 +2:
69 we can construct two palindrome with the 2 'a's:
70 'a' and 'aa'
71 note that since there is no 'a' in S[i+1...j-1],
72 so 'a' and 'aa' are not duplicate
73 */
74 dp[i][j] = dp[i+1][j-1]*2+2;
75 }else if(lo == hi){
76 /*
77 axxaxxa
78 dp[i+1][j-1]*2:
79 both the palindrome count in xxaxx itself and
80 axxaxxa, so two times
81
82 +1:
83 we can construct the palindrome 'aa'
84 with S[i] and S[j]
85 note that palindrome 'a' will be duplicate since
86 S[i+1...j-1] already contains one 'a'
87 */
88 dp[i][j] = dp[i+1][j-1]*2+1;
89 }else if(lo < hi){
90 /*
91 axxayyaxxa
92
93 dp[i+1][j-1]*2:
94 both the palindrome count in xxayyaxx itself and
95 axxayyaxxa, so two times
96
97 -dp[lo+1][hi-1]:
98 when constructing palindrome using the two 'a's
99 (S[i] and S[j]),
100 we may use S[i],S[j] and the yy(S[lo+1...hi-1]),
101 but since the palindrome constructed from
102 S[i],S[j],S[lo+1...hi-1] will be the same of
103 that constructed from S[lo],S[hi],S[lo+1...hi-1]
104 so we need to subtract dp[lo+1][hi-1] to remove
105 the duplicate
106 */
107 dp[i][j] = dp[i+1][j-1]*2 - dp[lo+1][hi-1];
108 }
109 }
110
111 /*
112 in
113 "dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];"
114 or
115 "dp[i][j] = dp[i+1][j-1]*2 - dp[lo+1][hi-1];"
116
117 because dp[i][j-1], dp[i+1][j], dp[i+1][j-1] and
118 dp[lo+1][hi-1] are after mod operation
119 (they are not original value),
120 so the subtraction may lead to negative number
121 (the negative number doesn't come from integer overflow)
122 knowning the reason why negative number appears,
123 it's naturally to add the number MOD back!
124 */
125 dp[i][j] = (dp[i][j] < 0) ? dp[i][j] + MOD: dp[i][j] % MOD;
126 }
127 }
128
129 return dp[0][n-1];
130 }
131};
132
133//DP, O(N^2)
134//optimize the previous method, finding lo and hi in O(1) time
135//https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/109507/Java-96ms-DP-Solution-with-Detailed-Explanation/111365
136//Runtime: 120 ms, faster than 84.48% of C++ online submissions for Count Different Palindromic Subsequences.
137//Memory Usage: 36.7 MB, less than 100.00% of C++ online submissions for Count Different Palindromic Subsequences.
138//time: O(N^2), space: O(N^2)
139class Solution {
140public:
141 int countPalindromicSubsequences(string S) {
142 int n = S.size();
143 const int MOD = 1e9+7;
144 vector<vector<int>> dp(n, vector(n, 0));
145
146 //record the next right/left position of the same char
147 vector<int> nextRight(n), nextLeft(n);
148 //helper
149 vector<int> char2pos(4, -1);
150
151 for(int i = 0; i < n; i++){
152 /*
153 initial value of char2pos are all -1,
154 meaning that we haven't found S[i] in S[0...i-1]
155 */
156 nextLeft[i] = char2pos[S[i]-'a'];
157 /*
158 record the position of S[i],
159 this will be used in following iterations
160 */
161 char2pos[S[i]-'a'] = i;
162 }
163
164 fill(char2pos.begin(), char2pos.end(), n);
165 for(int i = n-1; i >= 0; i--){
166 nextRight[i] = char2pos[S[i]-'a'];
167 char2pos[S[i]-'a'] = i;
168 }
169
170 for(int dist = 0; dist < n; dist++){
171 for(int i = 0; i + dist < n; i++){
172 int j = i+dist;
173 if(i == j){
174 dp[i][j] = 1;
175 continue;
176 }
177
178 if(S[i] != S[j]){
179 dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];
180 }else{
181 //the position of S[i] in S[i+1...]
182 int lo = nextRight[i];
183 //the position of S[j] in S[...j-1]
184 int hi = nextLeft[j];
185 /*
186 note that lo and hi could be out of the range of
187 S[i+1...j-1], but this can be handled by lo > hi
188 */
189 if(lo > hi){
190 dp[i][j] = dp[i+1][j-1]*2+2;
191 }else if(lo == hi){
192 dp[i][j] = dp[i+1][j-1]*2+1;
193 }else if(lo < hi){
194 dp[i][j] = dp[i+1][j-1]*2 - dp[lo+1][hi-1];
195 }
196 }
197 dp[i][j] = (dp[i][j] < 0) ? dp[i][j] + MOD: dp[i][j] % MOD;
198 }
199 }
200
201 return dp[0][n-1];
202 }
203};Cost
Complexity
Time
O(N^3), space: O(N^2)
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.