Stack applications (expression evaluation, backtracking) in Data Structures Theory - Time & Space Complexity
Stacks help solve problems like checking expressions or exploring choices step-by-step.
We want to know how the time needed grows as the input or problem size grows.
Analyze the time complexity of the following code snippet.
function evaluateExpression(expr) {
let stack = [];
for (let char of expr) {
if (isOperand(char)) {
stack.push(char);
} else if (isOperator(char)) {
let b = stack.pop();
let a = stack.pop();
let result = applyOperator(a, b, char);
stack.push(result);
}
}
return stack.pop();
}
This code evaluates a simple expression using a stack by processing each character once.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: Looping through each character of the expression once.
- How many times: Exactly once per character, so as many times as the expression length.
As the expression gets longer, the number of steps grows directly with its length.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 10 steps |
| 100 | About 100 steps |
| 1000 | About 1000 steps |
Pattern observation: Doubling the input roughly doubles the work needed.
Time Complexity: O(n)
This means the time to evaluate grows in a straight line with the size of the expression.
[X] Wrong: "Because there are nested operations, the time must be quadratic or worse."
[OK] Correct: Each character is processed once; the stack operations are quick and do not multiply the work.
Understanding how stacks handle tasks like expression evaluation or backtracking shows you can manage step-by-step problem solving efficiently.
"What if the expression included nested parentheses requiring recursive evaluation? How would the time complexity change?"