Evaluating Postfix Expressions Using a Stack — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Stack and solve the Evaluating Postfix Expressions Using a Stack problem optimally.
O(n)O(n)Problem Description
You are given a string that represents a postfix (Reverse Polish Notation) arithmetic expression. The expression contains only single‑digit operands (0–9) and the four basic operators: addition (+), subtraction (−), multiplication (×), and integer division (÷). The expression is guaranteed to be syntactically correct and to contain no division by zero. Your task is to evaluate the expression using a stack and return the final integer result as a string. Division should truncate toward zero, following the standard integer division rules of most programming languages.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Evaluating Postfix Expressions Using a Stack"
WHY DOES IT MATTER?
The stack pattern guarantees that each operator sees the correct operands without backtracking, which is essential for linear‑time evaluation of expressions that can be arbitrarily nested.
OPTIMIZATION CHALLENGE
Recognizing that each token is processed once and that the stack depth never exceeds the number of operands reduces both time to O(n) and space to O(n).
REAL-WORLD CONNECTION
It is analogous to how a CPU’s instruction stack processes function calls: arguments are pushed, the function executes, and the result is pushed back, enabling efficient call/return semantics.
During interviews, emphasize the stack’s role as an implicit parse tree and explain how popping two operands corresponds to evaluating a binary node.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
Postfix (Reverse Polish Notation) expressions eliminate the need for parentheses by placing operators after their operands. The evaluation algorithm processes the expression left‑to‑right, pushing operands onto a stack. When an operator is encountered, the algorithm pops the required number of operands (two for binary operators), applies the operation, and pushes the result back onto the stack. This guarantees that each subexpression is evaluated exactly once and that the final stack contains the overall result.
Naive approaches, such as recursively parsing the expression or repeatedly scanning the string to find subexpressions, suffer from exponential time or quadratic space due to repeated work and deep recursion. They also struggle with large inputs because each recursive call adds overhead and can lead to stack overflow. The stack‑based method, by contrast, runs in linear time and uses linear space proportional to the maximum depth of nested operations, making it optimal for long expressions.
The underlying algorithmic pattern is a depth‑first traversal of an implicit expression tree using an explicit stack. This pattern is a cornerstone of compiler design, expression evaluation, and many parsing problems. It transforms a linear token stream into a structured computation without constructing the entire parse tree, thereby achieving both time and space efficiency.
Interview Questions on This Problem
Q1How would you modify the algorithm to support multi‑digit operands and whitespace separators?
First, tokenize the input string by splitting on whitespace or using a scanner that recognizes numbers. Then, push each numeric token onto the stack as an integer. Operators are processed the same way, popping operands and pushing the result. This change preserves linear time and space while handling arbitrary integer sizes.
Q2What changes are required if the division operator should perform true floating‑point division instead of integer division?
Use a floating‑point data type (e.g., double) for the stack elements. When parsing operands, convert the string to a double. For the division operator, perform a / b using floating‑point arithmetic. Ensure that the stack holds doubles throughout to avoid truncation.
Q3In a distributed system, how could you evaluate a postfix expression that is split across multiple nodes?
Treat each node as a worker that evaluates a sub‑expression and returns its result. Use a tree‑like aggregation where parent nodes combine child results using the operators. This mirrors the stack approach but distributes the stack across nodes, requiring careful synchronization to maintain operand order.
Examples
Input
23+
Output
5
Explanation: Push 2, push 3, apply '+': 2 + 3 = 5. Result is 5.
Input
84/2-
Output
0
Explanation: Push 8, push 4, apply '/': 8 / 4 = 2. Push 2, apply '-': 2 - 2 = 0. Result is 0.
Input
93-42*+
Output
14
Explanation: Push 9, push 3, apply '-': 9 - 3 = 6. Push 4, push 2, apply '*': 4 * 2 = 8. Apply '+': 6 + 8 = 14. Result is 14.
Input
12+34+*
Output
21
Explanation: Push 1, push 2, apply '+': 1 + 2 = 3. Push 3, push 4, apply '+': 3 + 4 = 7. Apply '*': 3 * 7 = 21. Result is 21.
Constraints
- 1 <= expression.length <= 100000
- expression contains only characters '0'–'9', '+', '-', '*', '/'
- the expression is a valid postfix expression with no division by zero
- all intermediate and final results fit within a 32‑bit signed integer
Optimal Approach & Strategy
Use a stack: push operands, pop two for each operator, compute, and push the result. This processes each character once, yielding linear time and space.
Brute Force Approach
A naive method would repeatedly scan the string to find the first operator, evaluate the two preceding operands, replace them with the result, and repeat until one number remains. This approach has quadratic time complexity and is impractical for long expressions.
Code Solutions
const fs = require('fs');
const expr = fs.readFileSync(0, 'utf8').trim();
const stack = [];
for (const ch of expr) {
if (ch >= '0' && ch <= '9') {
stack.push(parseInt(ch, 10));
} else {
const b = stack.pop();
const a = stack.pop();
let res;
switch (ch) {
case '+': res = a + b; break;
case '-': res = a - b; break;
case '*': res = a * b; break;
case '/': res = Math.trunc(a / b); break; // integer division
}
stack.push(res);
}
}
console.log(stack[stack.length - 1]);#include <bits/stdc++.h>
using namespace std;
int main() {
string expr;
if (!(cin >> expr)) return 0;
vector<int> stack;
for (char c : expr) {
if (isdigit(c)) {
stack.push_back(c - '0');
} else {
int b = stack.back(); stack.pop_back();
int a = stack.back(); stack.pop_back();
int res = 0;
switch (c) {
case '+': res = a + b; break;
case '-': res = a - b; break;
case '*': res = a * b; break;
case '/': res = a / b; break; // integer division
}
stack.push_back(res);
}
}
cout << stack.back() << endl;
return 0;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
if (!sc.hasNext()) return;
String expr = sc.next();
Deque<Integer> stack = new ArrayDeque<>();
for (char ch : expr.toCharArray()) {
if (Character.isDigit(ch)) {
stack.push(ch - '0');
} else {
int b = stack.pop();
int a = stack.pop();
int res;
switch (ch) {
case '+': res = a + b; break;
case '-': res = a - b; break;
case '*': res = a * b; break;
case '/': res = a / b; break; // integer division
default: res = 0; // should not happen
}
stack.push(res);
}
}
System.out.println(stack.pop());
}
}
import sys
def evaluate(expr: str) -> int:
stack = []
for ch in expr:
if ch.isdigit():
stack.append(int(ch))
else:
b = stack.pop()
a = stack.pop()
if ch == '+':
stack.append(a + b)
elif ch == '-':
stack.append(a - b)
elif ch == '*':
stack.append(a * b)
elif ch == '/':
stack.append(int(a / b)) # truncates toward zero
return stack[-1]
if __name__ == "__main__":
expr = sys.stdin.read().strip()
print(evaluate(expr))
const fs = require('fs');
const expr = fs.readFileSync(0, 'utf8').trim();
const stack = [];
for (const ch of expr) {
if (ch >= '0' && ch <= '9') {
stack.push(parseInt(ch, 10));
} else {
const b = stack.pop();
const a = stack.pop();
let res;
switch (ch) {
case '+': res = a + b; break;
case '-': res = a - b; break;
case '*': res = a * b; break;
case '/': res = Math.trunc(a / b); break; // integer division
}
stack.push(res);
}
}
console.log(stack[stack.length - 1]);
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