This problem looks busy at first, but the accepted solution is built around one steady invariant. For 808. Soup Servings, the solution in this repository is mainly a dynamic programming 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: dynamic programming.
The notes already sitting in the source point us in the right direction:
- 20 / 41 test cases passed.
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 soupServings, prob, f.
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 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:
- 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 Error: stack-overflow on address
02//20 / 41 test cases passed.
03class Solution {
04public:
05 vector<int> serveAs;
06 unordered_map<string, vector<double>> memo;
07
08 vector<double> soupServings(int NA, int NB){
09 /*
10 prob[0]: A will be empty first
11 prob[1]: A and B become empty at the same time
12 */
13 string key = to_string(NA) + "_" + to_string(NB);
14 if(memo.find(key) != memo.end()){
15 return memo[key];
16 }
17
18 vector<double> prob(2);
19
20 for(int serveA : serveAs){
21 int serveB = 100 - serveA;
22 if(NA <= serveA && NB <= serveB){
23 prob[1] += 0.25;
24 }else if(NA <= serveA){
25 prob[0] += 0.25;
26 }else if(NB <= serveB){
27 //pass
28 }else{
29 vector<double> subprob = soupServings(NA-serveA, NB-serveB);
30 prob[0] += 0.25 * subprob[0];
31 prob[1] += 0.25 * subprob[1];
32 }
33 }
34
35 return memo[key] = prob;
36 };
37
38 double soupServings(int N) {
39 serveAs = {100, 75, 50, 25};
40 vector<double> prob = soupServings(N, N);
41 return prob[0] + prob[1] * 0.5;
42 }
43};
44
45//return 1 when N >= 4800
46/*
47Answers within 10^-5 of the true value will be accepted as correct.
48When N = 4800, the result = 0.999994994426
49*/
50//https://leetcode.com/problems/soup-servings/discuss/121711/C%2B%2BJavaPython-When-N-greater-4800-just-return-1
51//Runtime: 32 ms, faster than 15.42% of C++ online submissions for Soup Servings.
52//Memory Usage: 8.9 MB, less than 41.48% of C++ online submissions for Soup Servings.
53class Solution {
54public:
55 vector<int> serveAs;
56 unordered_map<string, vector<double>> memo;
57
58 vector<double> soupServings(int NA, int NB){
59 /*
60 prob[0]: A will be empty first
61 prob[1]: A and B become empty at the same time
62 */
63 if(NA >= 4800) return {1.0, 0.0};
64 string key = to_string(NA) + "_" + to_string(NB);
65 if(memo.find(key) != memo.end()){
66 return memo[key];
67 }
68
69 vector<double> prob(2);
70
71 for(int serveA : serveAs){
72 int serveB = 100 - serveA;
73 if(NA <= serveA && NB <= serveB){
74 prob[1] += 0.25;
75 }else if(NA <= serveA){
76 prob[0] += 0.25;
77 }else if(NB <= serveB){
78 //pass
79 }else{
80 vector<double> subprob = soupServings(NA-serveA, NB-serveB);
81 prob[0] += 0.25 * subprob[0];
82 prob[1] += 0.25 * subprob[1];
83 }
84 }
85
86 return memo[key] = prob;
87 };
88
89 double soupServings(int N) {
90 serveAs = {100, 75, 50, 25};
91 vector<double> prob = soupServings(N, N);
92 return prob[0] + prob[1] * 0.5;
93 }
94};
95
96//conversion
97//Runtime: 8 ms, faster than 52.57% of C++ online submissions for Soup Servings.
98//Memory Usage: 14.5 MB, less than 13.63% of C++ online submissions for Soup Servings.
99class Solution {
100public:
101 vector<vector<double>> memo;
102
103 double f(int NA, int NB){
104 if(NA <= 0 && NB <= 0) return 0.5;
105 if(NA <= 0) return 1.0;
106 if(NB <= 0) return 0.0;
107 if(memo[NA][NB] > 0) return memo[NA][NB];
108 return memo[NA][NB] = 0.25*(f(NA-4,NB)+f(NA-3,NB-1)+f(NA-2,NB-2)+f(NA-1,NB-3));
109 };
110
111 double soupServings(int N) {
112 if(N > 4800) return 1.0;
113 //200 because 4800/25+1=193, it's around 200
114 memo = vector<vector<double>>(200, vector<double>(200));
115 //conversion from 25's multiple to the number of spoons
116 return f(ceil((double)N/25), ceil((double)N/25));
117 }
118};
119
120//conversion, using array
121//https://leetcode.com/problems/soup-servings/discuss/121711/C%2B%2BJavaPython-When-N-greater-4800-just-return-1
122//Runtime: 4 ms, faster than 69.96% of C++ online submissions for Soup Servings.
123//Memory Usage: 6.2 MB, less than 95.45% of C++ online submissions for Soup Servings.
124class Solution {
125public:
126 double memo[200][200];
127
128 double f(int NA, int NB){
129 if(NA <= 0 && NB <= 0) return 0.5;
130 if(NA <= 0) return 1.0;
131 if(NB <= 0) return 0.0;
132 if(memo[NA][NB] > 0) return memo[NA][NB];
133 return memo[NA][NB] = 0.25*(f(NA-4,NB)+f(NA-3,NB-1)+f(NA-2,NB-2)+f(NA-1,NB-3));
134 };
135
136 double soupServings(int N) {
137 if(N > 4800) return 1.0;
138 //conversion from 25's multiple to the number of spoons
139 return f(ceil((double)N/25), ceil((double)N/25));
140 }
141};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.