Problem Understanding
Given numbers 1, 2, 3, ..., n that must be pushed onto a stack in order, generate all possible output sequences from interleaved push/pop operations.
Key Constraint: Push order is fixed, but pop timing is flexible.
Core Insight: Decision Tree Approach
At any point during the process, you have exactly two choices:
- Pop from stack (if not empty)
- Push next number (if available)
This creates a binary decision tree where each path represents a valid sequence.
Recursive State Definition
The recursive function tracks three pieces of state:
-
nextPush
: next number to push (1 to n) -
stack
: current stack contents -
output
: numbers popped so far
void generate(int nextPush, stack<int>& stk, vector<int>& output) {
if (output.size() == n) {
// Base case: complete sequence found
print(output);
return;
}
// Two recursive choices...
}
Step-by-Step Execution for n=2
Initial State: nextPush=1, stack=[], output=[]
Path 1: Push then Pop
Push 1 → stack=[1], nextPush=2
Push 2 → stack=[1,2], nextPush=3
Pop 2 → stack=[1], output=[2]
Pop 1 → stack=[], output=[2,1] ✓
Path 2: Interleaved Push/Pop
Push 1 → stack=[1], nextPush=2
Pop 1 → stack=[], output=[1]
Push 2 → stack=[2], nextPush=3
Pop 2 → stack=[], output=[1,2] ✓
Backtracking Implementation
void dfs(int n, int index, stack<int>& stk, vector<int>& path) {
if (path.size() == n) {
for (int num : path) cout << num;
cout << endl;
return;
}
// Choice 1: Pop from stack
if (!stk.empty()) {
int top_val = stk.top();
stk.pop();
path.push_back(top_val);
dfs(n, index, stk, path);
path.pop_back(); // Backtrack
stk.push(top_val);
}
// Choice 2: Push next number
if (index <= n) {
stk.push(index);
dfs(n, index + 1, stk, path);
stk.pop(); // Backtrack
}
}
Key Observations
- Catalan Numbers: The number of valid sequences equals the nth Catalan number
- Parentheses Connection: Push = '(', Pop = ')' - valid sequences correspond to balanced parentheses
- Tree Traversal: Similar to generating all possible binary tree traversals
Thinking Process Summary
- Identify the decision points (push vs pop)
- Define the recursive state
- Handle base case (complete output)
- Explore both choices with backtracking
- Recognize the combinatorial pattern
This approach transforms a complex combinatorial problem into a systematic exploration of a decision tree, demonstrating how recursive thinking can simplify seemingly difficult problems.
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