LeetCode #241 — MEDIUM

Different Ways to Add Parentheses

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.

The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 10⁴.

Example 1:

Input: expression = "2-1-1"
Output: [0,2]
Explanation:
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2:

Input: expression = "2*3-4*5"
Output: [-34,-14,-10,-10,10]
Explanation:
(2*(3-(4*5))) = -34 
((2*3)-(4*5)) = -14 
((2*(3-4))*5) = -10 
(2*((3-4)*5)) = -10 
(((2*3)-4)*5) = 10

Constraints:

1 <= expression.length <= 20
expression consists of digits and the operator '+', '-', and '*'.
All the integer values in the input expression are in the range [0, 99].
The integer values in the input expression do not have a leading '-' or '+' denoting the sign.

Step 01

Brute Force Baseline

Problem summary: Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order. The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 104.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

"2-1-1"

Example 2

"2*3-4*5"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #241: Different Ways to Add Parentheses
class Solution {
    private static Map<String, List<Integer>> memo = new HashMap<>();

    public List<Integer> diffWaysToCompute(String expression) {
        return dfs(expression);
    }

    private List<Integer> dfs(String exp) {
        if (memo.containsKey(exp)) {
            return memo.get(exp);
        }
        List<Integer> ans = new ArrayList<>();
        if (exp.length() < 3) {
            ans.add(Integer.parseInt(exp));
            return ans;
        }
        for (int i = 0; i < exp.length(); ++i) {
            char c = exp.charAt(i);
            if (c == '-' || c == '+' || c == '*') {
                List<Integer> left = dfs(exp.substring(0, i));
                List<Integer> right = dfs(exp.substring(i + 1));
                for (int a : left) {
                    for (int b : right) {
                        if (c == '-') {
                            ans.add(a - b);
                        } else if (c == '+') {
                            ans.add(a + b);
                        } else {
                            ans.add(a * b);
                        }
                    }
                }
            }
        }
        memo.put(exp, ans);
        return ans;
    }
}

// Accepted solution for LeetCode #241: Different Ways to Add Parentheses
var memo = map[string][]int{}

func diffWaysToCompute(expression string) []int {
	return dfs(expression)
}

func dfs(exp string) []int {
	if v, ok := memo[exp]; ok {
		return v
	}
	if len(exp) < 3 {
		v, _ := strconv.Atoi(exp)
		return []int{v}
	}
	ans := []int{}
	for i, c := range exp {
		if c == '-' || c == '+' || c == '*' {
			left, right := dfs(exp[:i]), dfs(exp[i+1:])
			for _, a := range left {
				for _, b := range right {
					if c == '-' {
						ans = append(ans, a-b)
					} else if c == '+' {
						ans = append(ans, a+b)
					} else {
						ans = append(ans, a*b)
					}
				}
			}
		}
	}
	memo[exp] = ans
	return ans
}

# Accepted solution for LeetCode #241: Different Ways to Add Parentheses
class Solution:
    def diffWaysToCompute(self, expression: str) -> List[int]:
        @cache
        def dfs(exp):
            if exp.isdigit():
                return [int(exp)]
            ans = []
            for i, c in enumerate(exp):
                if c in '-+*':
                    left, right = dfs(exp[:i]), dfs(exp[i + 1 :])
                    for a in left:
                        for b in right:
                            if c == '-':
                                ans.append(a - b)
                            elif c == '+':
                                ans.append(a + b)
                            else:
                                ans.append(a * b)
            return ans

        return dfs(expression)

// Accepted solution for LeetCode #241: Different Ways to Add Parentheses
struct Solution;

enum Tok {
    Op(char),
    Val(i32),
}

impl Solution {
    fn diff_ways_to_compute(input: String) -> Vec<i32> {
        let mut val = 0;
        let mut toks: Vec<Tok> = vec![];
        for c in input.chars() {
            match c {
                '+' | '-' | '*' => {
                    toks.push(Tok::Val(val));
                    toks.push(Tok::Op(c));
                    val = 0;
                }
                _ => {
                    val *= 10;
                    val += (c as u8 - b'0') as i32;
                }
            }
        }
        toks.push(Tok::Val(val));
        Self::eval(&toks)
    }

    fn eval(toks: &[Tok]) -> Vec<i32> {
        let mut res = vec![];
        let n = toks.len();
        for i in 0..n {
            if let Tok::Op(c) = toks[i] {
                let left = Self::eval(&toks[0..i]);
                let right = Self::eval(&toks[i + 1..n]);
                for l in &left {
                    for r in &right {
                        match c {
                            '+' => {
                                res.push(l + r);
                            }
                            '-' => {
                                res.push(l - r);
                            }
                            '*' => {
                                res.push(l * r);
                            }
                            _ => {}
                        }
                    }
                }
            }
        }
        if res.is_empty() {
            if let Tok::Val(val) = toks[0] {
                vec![val]
            } else {
                unreachable!()
            }
        } else {
            res
        }
    }
}

#[test]
fn test() {
    let input = "2-1-1".to_string();
    let mut res = vec![0, 2];
    let mut ans = Solution::diff_ways_to_compute(input);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
    let input = "2*3-4*5".to_string();
    let mut res = vec![-34, -14, -10, -10, 10];
    let mut ans = Solution::diff_ways_to_compute(input);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #241: Different Ways to Add Parentheses
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #241: Different Ways to Add Parentheses
// class Solution {
//     private static Map<String, List<Integer>> memo = new HashMap<>();
// 
//     public List<Integer> diffWaysToCompute(String expression) {
//         return dfs(expression);
//     }
// 
//     private List<Integer> dfs(String exp) {
//         if (memo.containsKey(exp)) {
//             return memo.get(exp);
//         }
//         List<Integer> ans = new ArrayList<>();
//         if (exp.length() < 3) {
//             ans.add(Integer.parseInt(exp));
//             return ans;
//         }
//         for (int i = 0; i < exp.length(); ++i) {
//             char c = exp.charAt(i);
//             if (c == '-' || c == '+' || c == '*') {
//                 List<Integer> left = dfs(exp.substring(0, i));
//                 List<Integer> right = dfs(exp.substring(i + 1));
//                 for (int a : left) {
//                     for (int b : right) {
//                         if (c == '-') {
//                             ans.add(a - b);
//                         } else if (c == '+') {
//                             ans.add(a + b);
//                         } else {
//                             ans.add(a * b);
//                         }
//                     }
//                 }
//             }
//         }
//         memo.put(exp, ans);
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.