The trick here is to name the state correctly, then let the implementation follow. For 42. Trapping Rain Water, 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, stack, sliding window.
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 trap, leftMax, tmp.
Guide
Why?
The solution works because it narrows the problem until every update has a clear reason to exist.
- The stack stores unfinished context, which is usually the cleanest way to handle nested or monotonic structure.
- 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//Runtime: 4 ms, faster than 95.59% of C++ online submissions for Trapping Rain Water.
02//Memory Usage: 8.2 MB, less than 100.00% of C++ online submissions for Trapping Rain Water.
03class Solution {
04public:
05 int trap(vector<int>& height) {
06 int left = 0, right = 0;
07 int ans = 0;
08 int N = height.size();
09
10 while(right < N){
11 int valley = height[left];
12
13 //left wall should have height > 0
14 while(left < N && height[left] == 0){
15 left++;
16 }
17 //cannot find left wall
18 if(left == N) break;
19
20 //right wall should be larger than valley
21 for(right = left+1; right < N && height[right] <= valley; right++){
22 valley = min(valley, height[right]);
23 }
24 //cannot find right wall
25 if(right == N) break;
26 // cout << "left: " << left << ", right: " << right << endl;
27
28 //we've found a valid right wall,
29 //now we want to discover higher right wall
30 //we only need to find a higher right wall when right wall is lower than left wall
31 int head = right+1;
32 if(height[left] > height[right]){
33 int last = right;
34 while(head < N && height[head] < height[left]){
35 if(height[head] > height[last]){
36 last = head;
37 }
38 head++;
39 }
40 if(head < N && height[head] >= height[left]){
41 //found a right wall >= left wall
42 right = head;
43 }else{
44 //found the highest right wall > original right wall
45 right = last;
46 }
47 }
48
49 //calculate current trapping water amount
50 int lh = height[left], rh = height[right];
51 for(int i = left+1; i < min(right, N); i++){
52 // cout << i << ", height: " << height[i] << endl;
53 //we should take max(0, x) here!
54 ans += max(0, min(lh, rh) - height[i]);
55 }
56 // cout << left << " " << right << " " << valley << " " << ans << endl;
57
58 left = right;
59 }
60
61 return ans;
62 }
63};
64
65//Approach 1: Brute force
66//Runtime: 312 ms, faster than 5.10% of C++ online submissions for Trapping Rain Water.
67//Memory Usage: 8.5 MB, less than 100.00% of C++ online submissions for Trapping Rain Water.
68//time: O(n^2), space: O(1)
69class Solution {
70public:
71 int trap(vector<int>& height) {
72 int ans = 0;
73
74 for(int i = 0; i < height.size(); i++){
75 int maxLeft = 0, maxRight = 0;
76 for(int j = i; j >= 0 ; j--){
77 maxLeft = max(maxLeft, height[j]);
78 }
79 for(int j = i; j < height.size() ; j++){
80 maxRight = max(maxRight, height[j]);
81 }
82 ans += min(maxLeft, maxRight) - height[i];
83 // cout << maxLeft << " " << maxRight << " " << ans << endl;
84 }
85
86 return ans;
87 }
88};
89
90//Approach 2: Dynamic Programming
91//Runtime: 4 ms, faster than 95.59% of C++ online submissions for Trapping Rain Water.
92//Memory Usage: 8.2 MB, less than 100.00% of C++ online submissions for Trapping Rain Water.
93//time: O(n), space: O(n)
94class Solution {
95public:
96 int trap(vector<int>& height) {
97 int N = height.size();
98 if(N == 0) return 0;
99 int ans = 0;
100
101 vector<int> leftMax(N), rightMax(N);
102 leftMax[0] = height[0];
103 for(int i = 1; i < N; i++){
104 leftMax[i] = max(leftMax[i-1], height[i]);
105 }
106
107 rightMax[N-1] = height[N-1];
108 for(int i = N-2; i >= 0; i--){
109 rightMax[i] = max(rightMax[i+1], height[i]);
110 }
111
112 for(int i = 1; i < N-1; i++){
113 ans += min(leftMax[i], rightMax[i]) - height[i];
114 }
115
116 return ans;
117 }
118};
119
120//Approach 3: Using stacks
121//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Trapping Rain Water.
122//Memory Usage: 8.5 MB, less than 100.00% of C++ online submissions for Trapping Rain Water.
123//time: O(n), space: O(n)
124class Solution {
125public:
126 int trap(vector<int>& height) {
127 int ans = 0, cur = 0;
128 stack<int> stk;
129 for(int cur = 0; cur < height.size(); cur++){
130 // cout << cur << endl;
131 // if(!stk.empty()){
132 // cout << "stack: ";
133 // vector<int> tmp(&stk.top()+1-stk.size(), &stk.top()+1);
134 // for(int i = 0; i < tmp.size(); i++){
135 // cout << tmp[i] << " ";
136 // }
137 // cout << endl;
138 // }
139 while(!stk.empty() && height[cur] > height[stk.top()]){
140 int top = stk.top(); stk.pop();
141 // cout << "pop: " << top << endl;
142 if(stk.empty()){
143 break;
144 }
145 //top is bounded by previous bar in the stack and current bar
146 int dist = cur - stk.top() - 1;
147 //the left and right walls are : stk.top(), cur
148 int bounded_height = min(height[stk.top()], height[cur]) - height[top];
149 ans += dist * bounded_height;
150 // cout << "[" << stk.top() << ", " << cur << "] " << ans << endl;
151 }
152 //the current bar is bounded by previous bar in the stack
153 stk.push(cur);
154 }
155 return ans;
156 }
157};
158
159//Approach 4: Using 2 pointers
160//Runtime: 4 ms, faster than 95.59% of C++ online submissions for Trapping Rain Water.
161//Memory Usage: 8.3 MB, less than 100.00% of C++ online submissions for Trapping Rain Water.
162//time: O(n), space: O(1)
163class Solution {
164public:
165 int trap(vector<int>& height) {
166 int N = height.size();
167 if(N == 0) return 0;
168 int left = 0, right = N-1;
169 int leftMax = 0, rightMax = 0;
170 int ans = 0;
171 while(left < right){
172 if(height[left] > leftMax){
173 leftMax = height[left];
174 }
175 if(height[right] > rightMax){
176 rightMax = height[right];
177 }
178 if(leftMax < rightMax){
179 //use leftMax to substract because leftMax is min(leftMax, rightMax)
180 ans += max(0, leftMax - height[left]);
181 // cout << "[" << left << ", " << right << "] " << leftMax << " " << rightMax << " " << ans << endl;
182 left++;
183 }else{
184 ans += max(0, rightMax - height[right]);
185 // cout << "[" << left << ", " << right << "] " << leftMax << " " << rightMax << " " << ans << endl;
186 right--;
187 }
188 }
189 return ans;
190 }
191};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.