LeetCode #282 — HARD

Expression Add Operators

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators '+', '-', and/or '*' between the digits of num so that the resultant expression evaluates to the target value.

Note that operands in the returned expressions should not contain leading zeros.

Note that a number can contain multiple digits.

Example 1:

Input: num = "123", target = 6
Output: ["1*2*3","1+2+3"]
Explanation: Both "1*2*3" and "1+2+3" evaluate to 6.

Example 2:

Input: num = "232", target = 8
Output: ["2*3+2","2+3*2"]
Explanation: Both "2*3+2" and "2+3*2" evaluate to 8.

Example 3:

Input: num = "3456237490", target = 9191
Output: []
Explanation: There are no expressions that can be created from "3456237490" to evaluate to 9191.

Constraints:

1 <= num.length <= 10
num consists of only digits.
-2³¹ <= target <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators '+', '-', and/or '*' between the digits of num so that the resultant expression evaluates to the target value. Note that operands in the returned expressions should not contain leading zeros. Note that a number can contain multiple digits.

Baseline thinking

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

Pattern signal: Math · Backtracking

Example 1

"123"
6

Example 2

"232"
8

Example 3

"3456237490"
9191

Core Insight

What unlocks the optimal approach

Note that a number can contain multiple digits.
Since the question asks us to find all of the valid expressions, we need a way to iterate over all of them. (Hint: Recursion!)
We can keep track of the expression string and evaluate it at the very end. But that would take a lot of time. Can we keep track of the expression's value as well so as to avoid the evaluation at the very end of recursion?
Think carefully about the multiply operator. It has a higher precedence than the addition and subtraction operators. 1 + 2 = 3 1 + 2 - 4 --> 3 - 4 --> -1 1 + 2 - 4 * 12 --> -1 * 12 --> -12 (WRONG!) 1 + 2 - 4 * 12 --> -1 - (-4) + (-4 * 12) --> 3 + (-48) --> -45 (CORRECT!)
We simply need to keep track of the last operand in our expression and reverse it's effect on the expression's value while considering the multiply operator.

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

Largest constraint values

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 #282: Expression Add Operators
class Solution {
    private List<String> ans;
    private String num;
    private int target;

    public List<String> addOperators(String num, int target) {
        ans = new ArrayList<>();
        this.num = num;
        this.target = target;
        dfs(0, 0, 0, "");
        return ans;
    }

    private void dfs(int u, long prev, long curr, String path) {
        if (u == num.length()) {
            if (curr == target) ans.add(path);
            return;
        }
        for (int i = u; i < num.length(); i++) {
            if (i != u && num.charAt(u) == '0') {
                break;
            }
            long next = Long.parseLong(num.substring(u, i + 1));
            if (u == 0) {
                dfs(i + 1, next, next, path + next);
            } else {
                dfs(i + 1, next, curr + next, path + "+" + next);
                dfs(i + 1, -next, curr - next, path + "-" + next);
                dfs(i + 1, prev * next, curr - prev + prev * next, path + "*" + next);
            }
        }
    }
}

// Accepted solution for LeetCode #282: Expression Add Operators
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #282: Expression Add Operators
// class Solution {
//     private List<String> ans;
//     private String num;
//     private int target;
// 
//     public List<String> addOperators(String num, int target) {
//         ans = new ArrayList<>();
//         this.num = num;
//         this.target = target;
//         dfs(0, 0, 0, "");
//         return ans;
//     }
// 
//     private void dfs(int u, long prev, long curr, String path) {
//         if (u == num.length()) {
//             if (curr == target) ans.add(path);
//             return;
//         }
//         for (int i = u; i < num.length(); i++) {
//             if (i != u && num.charAt(u) == '0') {
//                 break;
//             }
//             long next = Long.parseLong(num.substring(u, i + 1));
//             if (u == 0) {
//                 dfs(i + 1, next, next, path + next);
//             } else {
//                 dfs(i + 1, next, curr + next, path + "+" + next);
//                 dfs(i + 1, -next, curr - next, path + "-" + next);
//                 dfs(i + 1, prev * next, curr - prev + prev * next, path + "*" + next);
//             }
//         }
//     }
// }

# Accepted solution for LeetCode #282: Expression Add Operators
class Solution:
    def addOperators(self, num: str, target: int) -> List[str]:
        ans = []

        def dfs(u, prev, curr, path):
            if u == len(num):
                if curr == target:
                    ans.append(path)
                return
            for i in range(u, len(num)):
                if i != u and num[u] == '0':
                    break
                next = int(num[u : i + 1])
                if u == 0:
                    dfs(i + 1, next, next, path + str(next))
                else:
                    dfs(i + 1, next, curr + next, path + "+" + str(next))
                    dfs(i + 1, -next, curr - next, path + "-" + str(next))
                    dfs(
                        i + 1,
                        prev * next,
                        curr - prev + prev * next,
                        path + "*" + str(next),
                    )

        dfs(0, 0, 0, "")
        return ans

// Accepted solution for LeetCode #282: Expression Add Operators
struct Solution;

impl Solution {
    fn add_operators(num: String, target: i32) -> Vec<String> {
        let mut res = vec![];
        for i in 1..=num.len() {
            let expr = &num[0..i];
            let val = expr.parse::<i64>().unwrap();
            if expr.len() > 1 && &expr[0..1] == "0" {
                break;
            }
            Self::dfs(
                i,
                expr.to_string(),
                val,
                val,
                '#',
                &mut res,
                &num,
                target as i64,
            );
        }
        res
    }

    fn dfs(
        start: usize,
        expr: String,
        val: i64,
        prev_val: i64,
        prev_op: char,
        all: &mut Vec<String>,
        num: &str,
        target: i64,
    ) {
        if start == num.len() {
            if val == target {
                all.push(expr);
            }
        } else {
            for end in start + 1..=num.len() {
                let cur_expr = &num[start..end];
                let cur_val = cur_expr.parse::<i64>().unwrap();
                if cur_expr.len() > 1 && &cur_expr[0..1] == "0" {
                    break;
                }
                Self::dfs(
                    end,
                    format!("{}+{}", expr, cur_expr),
                    val + cur_val,
                    cur_val,
                    '+',
                    all,
                    num,
                    target,
                );
                Self::dfs(
                    end,
                    format!("{}-{}", expr, cur_expr),
                    val - cur_val,
                    cur_val,
                    '-',
                    all,
                    num,
                    target,
                );
                Self::dfs(
                    end,
                    format!("{}*{}", expr, cur_expr),
                    match prev_op {
                        '+' => val - prev_val + prev_val * cur_val,
                        '-' => val + prev_val - prev_val * cur_val,
                        _ => prev_val * cur_val,
                    },
                    prev_val * cur_val,
                    prev_op,
                    all,
                    num,
                    target,
                )
            }
        }
    }
}

#[test]
fn test() {
    let num = "123".to_string();
    let target = 6;
    let mut res = vec_string!["1+2+3", "1*2*3"];
    let mut ans = Solution::add_operators(num, target);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let num = "232".to_string();
    let target = 8;
    let mut res = vec_string!["2*3+2", "2+3*2"];
    let mut ans = Solution::add_operators(num, target);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let num = "105".to_string();
    let target = 5;
    let mut res = vec_string!["1*0+5", "10-5"];
    let mut ans = Solution::add_operators(num, target);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let num = "00".to_string();
    let target = 0;
    let mut res = vec_string!["0+0", "0-0", "0*0"];
    let mut ans = Solution::add_operators(num, target);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let num = "3456237490".to_string();
    let target = 9191;
    let mut res = vec_string![];
    let mut ans = Solution::add_operators(num, target);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #282: Expression Add Operators
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #282: Expression Add Operators
// class Solution {
//     private List<String> ans;
//     private String num;
//     private int target;
// 
//     public List<String> addOperators(String num, int target) {
//         ans = new ArrayList<>();
//         this.num = num;
//         this.target = target;
//         dfs(0, 0, 0, "");
//         return ans;
//     }
// 
//     private void dfs(int u, long prev, long curr, String path) {
//         if (u == num.length()) {
//             if (curr == target) ans.add(path);
//             return;
//         }
//         for (int i = u; i < num.length(); i++) {
//             if (i != u && num.charAt(u) == '0') {
//                 break;
//             }
//             long next = Long.parseLong(num.substring(u, i + 1));
//             if (u == 0) {
//                 dfs(i + 1, next, next, path + next);
//             } else {
//                 dfs(i + 1, next, curr + next, path + "+" + next);
//                 dfs(i + 1, -next, curr - next, path + "-" + next);
//                 dfs(i + 1, prev * next, curr - prev + prev * next, path + "*" + next);
//             }
//         }
//     }
// }

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!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

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.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.