The trick here is to name the state correctly, then let the implementation follow. For 403. Frog Jump, the solution in this repository is mainly a graph traversal 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: graph traversal, dynamic programming.
The notes already sitting in the source point us in the right direction:
- BFS, unordered_set
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 canCross.
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.
- A set is doing the membership or uniqueness work, which keeps the main loop readable.
- The queue gives the solution a level-by-level or frontier-style traversal.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
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//BFS, unordered_set
02//Runtime: 1836 ms, faster than 5.05% of C++ online submissions for Frog Jump.
03//Memory Usage: 42.3 MB, less than 34.48% of C++ online submissions for Frog Jump.
04/*
05if using ordinary set:
06//TLE
07//25 / 39 test cases passed.
08*/
09struct pair_hash {
10 inline std::size_t operator()(const std::pair<int,int> & v) const {
11 return v.first*31+v.second;
12 }
13};
14
15class Solution {
16public:
17 bool canCross(vector<int>& stones) {
18 int goal = stones.back();
19
20 unordered_map<int, int> pos2idx;
21 for(int i = 0; i < stones.size(); ++i){
22 pos2idx[stones[i]] = i;
23 }
24
25 queue<pair<int, int>> q;
26 pair<int, int> cur = {0, 1};
27 unordered_set<pair<int, int>, pair_hash> visited;
28
29 q.push(cur);
30 visited.insert(cur);
31
32 while(!q.empty()){
33 cur = q.front(); q.pop();
34 int pos = cur.first, step = cur.second;
35 // cout << pos << ", " << step << endl;
36 if(pos == goal) return true;
37
38 int npos = pos+step;
39 int nstep;
40
41 //next position must be a stone
42 if(find(stones.begin(), stones.end(), npos) == stones.end()){
43 continue;
44 }
45
46 for(int i = -1; i <= 1; ++i){
47 nstep = step+i;
48 if(nstep <= 0) continue;
49 if(visited.find({npos, nstep}) == visited.end()){
50 visited.insert({npos, nstep});
51 q.push({npos, nstep});
52 }
53 }
54 }
55
56 return false;
57 }
58};
59
60//unordered_map
61//https://leetcode.com/problems/frog-jump/discuss/88824/Very-easy-to-understand-JAVA-solution-with-explanations
62//Runtime: 392 ms, faster than 33.90% of C++ online submissions for Frog Jump.
63//Memory Usage: 44.3 MB, less than 29.73% of C++ online submissions for Frog Jump.
64class Solution {
65public:
66 bool canCross(vector<int>& stones) {
67 if(stones.size() == 0) return true;
68 set<int> sstones(stones.begin(), stones.end());
69 stones = vector<int>(sstones.begin(), sstones.end());
70
71 int goal = stones.back();
72 // cout << "goal: " << goal << endl;
73
74 unordered_map<int, unordered_set<int>> pos2steps;
75 //the first position is always 0
76 pos2steps[0].insert(1);
77
78 int n = stones.size();
79 for(int i = 0; i < n; ++i){
80 int pos = stones[i];
81 for(int step : pos2steps[pos]){
82 int npos = pos + step;
83 if(npos == goal) return true;
84 if(sstones.find(npos) == sstones.end()){
85 //there is no stone here
86 continue;
87 }
88 for(int nstep = step-1; nstep <= step+1; ++nstep){
89 if(nstep <= 0) continue;
90 pos2steps[npos].insert(nstep);
91 }
92 }
93 }
94
95 return false;
96 }
97};
98
99//DP
100//https://leetcode.com/problems/frog-jump/discuss/193816/Concise-and-fast-DP-solution-using-2D-array-instead-of-HashMap-with-text-and-video-explanation.
101//Runtime: 260 ms, faster than 52.89% of C++ online submissions for Frog Jump.
102//Memory Usage: 18.3 MB, less than 70.73% of C++ online submissions for Frog Jump.
103class Solution {
104public:
105 bool canCross(vector<int>& stones) {
106 if(stones.size() == 0) return true;
107 set<int> sstones(stones.begin(), stones.end());
108 stones = vector<int>(sstones.begin(), sstones.end());
109
110 int goal = stones.back();
111
112 int n = stones.size();
113
114 /*
115 dp[i][j]: at ith stone(i is for index, not position!),
116 (the index of i is 0-based),
117 whether we can jump with distance j
118 n+1: at position i, the maximum distance we can jump is i+1
119 */
120 vector<vector<bool>> dp(n, vector<bool>(n+1, false));
121
122 //base case: at position 0, we can jump 1 distance forward
123 dp[0][1] = true;
124
125 for(int end = 1; end < n; ++end){
126 for(int start = 0; start < end; ++start){
127 //start and end are index, here convert them into actual positions
128 int dist = stones[end] - stones[start];
129 if(dist >= n+1 || !dp[start][dist]) continue;
130 if(end == n-1) return true;
131 for(int ndist = dist-1; ndist <= dist+1; ndist++){
132 if(ndist <= 0) continue;
133 dp[end][ndist] = true;
134 }
135 }
136 }
137
138 return false;
139 }
140};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.