Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
A boolean expression is an expression that evaluates to either true or false. It can be in one of the following shapes:
't' that evaluates to true.'f' that evaluates to false.'!(subExpr)' that evaluates to the logical NOT of the inner expression subExpr.'&(subExpr1, subExpr2, ..., subExprn)' that evaluates to the logical AND of the inner expressions subExpr1, subExpr2, ..., subExprn where n >= 1.'|(subExpr1, subExpr2, ..., subExprn)' that evaluates to the logical OR of the inner expressions subExpr1, subExpr2, ..., subExprn where n >= 1.Given a string expression that represents a boolean expression, return the evaluation of that expression.
It is guaranteed that the given expression is valid and follows the given rules.
Example 1:
Input: expression = "&(|(f))" Output: false Explanation: First, evaluate |(f) --> f. The expression is now "&(f)". Then, evaluate &(f) --> f. The expression is now "f". Finally, return false.
Example 2:
Input: expression = "|(f,f,f,t)" Output: true Explanation: The evaluation of (false OR false OR false OR true) is true.
Example 3:
Input: expression = "!(&(f,t))" Output: true Explanation: First, evaluate &(f,t) --> (false AND true) --> false --> f. The expression is now "!(f)". Then, evaluate !(f) --> NOT false --> true. We return true.
Constraints:
1 <= expression.length <= 2 * 104'(', ')', '&', '|', '!', 't', 'f', and ','.Problem summary: A boolean expression is an expression that evaluates to either true or false. It can be in one of the following shapes: 't' that evaluates to true. 'f' that evaluates to false. '!(subExpr)' that evaluates to the logical NOT of the inner expression subExpr. '&(subExpr1, subExpr2, ..., subExprn)' that evaluates to the logical AND of the inner expressions subExpr1, subExpr2, ..., subExprn where n >= 1. '|(subExpr1, subExpr2, ..., subExprn)' that evaluates to the logical OR of the inner expressions subExpr1, subExpr2, ..., subExprn where n >= 1. Given a string expression that represents a boolean expression, return the evaluation of that expression. It is guaranteed that the given expression is valid and follows the given rules.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"&(|(f))"
"|(f,f,f,t)"
"!(&(f,t))"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1106: Parsing A Boolean Expression
class Solution {
public boolean parseBoolExpr(String expression) {
Deque<Character> stk = new ArrayDeque<>();
for (char c : expression.toCharArray()) {
if (c != '(' && c != ')' && c != ',') {
stk.push(c);
} else if (c == ')') {
int t = 0, f = 0;
while (stk.peek() == 't' || stk.peek() == 'f') {
t += stk.peek() == 't' ? 1 : 0;
f += stk.peek() == 'f' ? 1 : 0;
stk.pop();
}
char op = stk.pop();
c = 'f';
if ((op == '!' && f > 0) || (op == '&' && f == 0) || (op == '|' && t > 0)) {
c = 't';
}
stk.push(c);
}
}
return stk.peek() == 't';
}
}
// Accepted solution for LeetCode #1106: Parsing A Boolean Expression
func parseBoolExpr(expression string) bool {
stk := []rune{}
for _, c := range expression {
if c != '(' && c != ')' && c != ',' {
stk = append(stk, c)
} else if c == ')' {
var t, f int
for stk[len(stk)-1] == 't' || stk[len(stk)-1] == 'f' {
if stk[len(stk)-1] == 't' {
t++
} else {
f++
}
stk = stk[:len(stk)-1]
}
op := stk[len(stk)-1]
stk = stk[:len(stk)-1]
c = 'f'
if (op == '!' && f > 0) || (op == '&' && f == 0) || (op == '|' && t > 0) {
c = 't'
}
stk = append(stk, c)
}
}
return stk[0] == 't'
}
# Accepted solution for LeetCode #1106: Parsing A Boolean Expression
class Solution:
def parseBoolExpr(self, expression: str) -> bool:
stk = []
for c in expression:
if c in 'tf!&|':
stk.append(c)
elif c == ')':
t = f = 0
while stk[-1] in 'tf':
t += stk[-1] == 't'
f += stk[-1] == 'f'
stk.pop()
match stk.pop():
case '!':
c = 't' if f else 'f'
case '&':
c = 'f' if f else 't'
case '|':
c = 't' if t else 'f'
stk.append(c)
return stk[0] == 't'
// Accepted solution for LeetCode #1106: Parsing A Boolean Expression
impl Solution {
fn dfs(i: &mut usize, expr: &[u8]) -> Vec<bool> {
let n = expr.len();
let mut res = Vec::new();
while *i < n {
let c = expr[*i];
*i += 1;
match c {
b')' => {
break;
}
b't' => {
res.push(true);
}
b'f' => {
res.push(false);
}
b'!' => {
res.push(!Self::dfs(i, expr)[0]);
}
b'&' => {
res.push(Self::dfs(i, expr).iter().all(|v| *v));
}
b'|' => {
res.push(Self::dfs(i, expr).iter().any(|v| *v));
}
_ => {}
}
}
res
}
pub fn parse_bool_expr(expression: String) -> bool {
let expr = expression.as_bytes();
let mut i = 0;
Self::dfs(&mut i, expr)[0]
}
}
// Accepted solution for LeetCode #1106: Parsing A Boolean Expression
function parseBoolExpr(expression: string): boolean {
const expr = expression;
const n = expr.length;
let i = 0;
const dfs = () => {
let res: boolean[] = [];
while (i < n) {
const c = expr[i++];
if (c === ')') {
break;
}
if (c === '!') {
res.push(!dfs()[0]);
} else if (c === '|') {
res.push(dfs().some(v => v));
} else if (c === '&') {
res.push(dfs().every(v => v));
} else if (c === 't') {
res.push(true);
} else if (c === 'f') {
res.push(false);
}
}
return res;
};
return dfs()[0];
}
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.