This problem looks busy at first, but the accepted solution is built around one steady invariant. For 1255. Maximum Score Words Formed by Letters, the solution in this repository is mainly a straightforward implementation 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: straightforward implementation.
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are maxScoreWords, updateScore, generateSubset.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- A map keeps the lookup side cheap; the code pays a little memory to avoid repeated searching.
- 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:
- Start from the smallest reliable state.
- Expand one legal move at a time.
- Cache, count, or merge information as soon as it becomes settled.
- Let the final stored value answer the original question.
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//Runtime: 292 ms, faster than 7.68% of C++ online submissions for Maximum Score Words Formed by Letters.
02//Memory Usage: 53.4 MB, less than 100.00% of C++ online submissions for Maximum Score Words Formed by Letters.
03
04class Solution {
05public:
06 int maxScoreWords(vector<string>& words, vector<char>& letters, vector<int>& score) {
07 int wordScore, combScore, maxScore = 0;
08 vector<int> wordsScore;
09 map<char, int> lettersCount, wordCount, combCount;
10 bool combExist;
11
12 //create the map lettersCount
13 for(char c : letters){
14 if(lettersCount.find(c) == lettersCount.end()){
15 lettersCount[c] = 1;
16 }else{
17 lettersCount[c]++;
18 }
19 }
20
21 //calculate scores of words, fill wordsScore
22 for(string word : words){
23 wordCount.clear();
24 wordScore = 0;
25 for(char c : word){
26 if(wordCount.find(c) == wordCount.end()){
27 wordCount[c] = 1;
28 }else{
29 wordCount[c]++;
30 }
31
32 if(wordCount[c] <= lettersCount[c]){
33 wordScore += score[c-'a'];
34 }else{
35 //this word cannot be formed
36 wordScore = 0;
37 break;
38 }
39 }
40 wordsScore.push_back(wordScore);
41 // cout << wordScore << " ";
42 }
43 // cout << endl;
44
45 //enumerate 2^words.length possibles
46 for(int num = 0; num < pow(2, words.size()); num++){
47 combScore = 0;
48 combCount.clear();
49 combExist = true;
50 for(int i = 0; i < words.size(); i++){
51 //(num/(pow(2,i)))%2: imagine "num" as a binary number
52 //then num/(pow(2,i)) is the part that after "num"'s i'th bit
53 //(num/(pow(2,i)))%2 is the i'th bit of "num"
54 if( (int)(num/(pow(2,i))) %2 == 1){
55 //check if this combination exist
56 for(char c : words[i]){
57 if(combCount.find(c) == combCount.end()){
58 combCount[c] = 1;
59 }else{
60 combCount[c]++;
61 }
62
63 if(combCount[c] > lettersCount[c]){
64 combExist = false;
65 }
66 }
67
68 if(!combExist){
69 combScore = 0;
70 break;
71 }else{
72 combScore += wordsScore[i];
73 }
74 }
75 }
76
77 if(combScore > maxScore){
78 // for(int i = 0; i < words.size(); i++){
79 // cout << (int)(num/(pow(2,i))) %2;
80 // }
81 // cout << endl;
82
83 maxScore = combScore;
84 }
85 }
86
87 return maxScore;
88 }
89};
90
91//Runtime: 48 ms, faster than 20.85% of C++ online submissions for Maximum Score Words Formed by Letters.
92//Memory Usage: 36.2 MB, less than 100.00% of C++ online submissions for Maximum Score Words Formed by Letters.
93//https://leetcode.com/problems/maximum-score-words-formed-by-letters/discuss/425104/Detailed-Explanation-using-Recursion
94class Solution {
95public:
96 int maxScore = INT_MIN;
97 vector<bool> taken;
98 vector<int> count;
99 //same as the argument of maxScoreWords
100 vector<string> words;
101 vector<int> score;
102
103 void updateScore(){
104 int curScore = 0;
105 vector<int> curCount = vector<int>(26, 0);
106 for(int i = 0; i < words.size(); i++){
107 if(taken[i]){
108 for(char c : words[i]){
109 // cout << "curCount: " << curCount[c-'a'] << endl;
110 // cout << "count: " << count[c-'a'] << endl;
111 // cout << "curScore: " << curScore << endl;
112 curCount[c-'a']++;
113 if(curCount[c-'a'] > count[c-'a']){
114 return;
115 }
116 curScore += score[c-'a'];
117 }
118 }
119 }
120 maxScore = max(maxScore, curScore);
121 };
122
123 void generateSubset(int n){
124 //every time a subset("taken") is generated,
125 //call updateScore to calculate the score of current combination,
126 //and update maxScore
127 if(n == 0){
128 updateScore();
129 return;
130 }
131
132 taken[n-1] = true;
133 generateSubset(n-1);
134
135 taken[n-1] = false;
136 generateSubset(n-1);
137 };
138
139 int maxScoreWords(vector<string>& words, vector<char>& letters, vector<int>& score) {
140 taken = vector<bool>(words.size(), false);
141 count = vector<int>(26, 0);
142 for(char c : letters){
143 count[c-'a']++;
144 }
145 this->words = words;
146 this->score = score;
147
148 generateSubset(words.size());
149 return maxScore;
150 }
151};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.