How do you solve Expression Add Operators on LeetCode?

Expression Add Operators (LeetCode #282) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(4^n) time complexity and O(n) space complexity. Key insight: Understanding operator precedence is crucial when building expressions.

What is the brute force solution for Expression Add Operators?

The brute force approach for Expression Add Operators is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries every possible combination of inserting operators between the digits of the input string. This means we will generate all possible expressions and evaluate them to see if they match the target. The optimal Optimal Solution approach improves this to O(4^n).

What algorithm or data structure is used to solve Expression Add Operators?

Expression Add Operators uses the following concepts: Math, String, Backtracking. The recommended patterns to study are: Backtracking, Recursion.

What are the interview tips for Expression Add Operators?

Explain your thought process clearly while coding. Discuss edge cases and how you would handle them. Practice writing recursive functions, as they are common in backtracking problems.

What are common mistakes when solving Expression Add Operators?

Forgetting to handle leading zeros in multi-digit numbers. Not considering operator precedence during evaluation.

#282

Expression Add Operators

Hard

MathStringBacktrackingBacktrackingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(4^n)
Space	O(1)	O(n)

💡

Intuition

Time O(4^n)Space O(n)

The optimal solution uses backtracking to explore possible expressions recursively. This method allows us to build expressions dynamically and evaluate them on-the-fly, avoiding the need to generate all combinations upfront.

⚙️

Algorithm

3 steps

1Step 1: Use a recursive function to build expressions by adding digits and operators.
2Step 2: Keep track of the current value and the last operand to handle multiplication correctly.
3Step 3: If the end of the string is reached, check if the current value matches the target.

solution.py19 lines

1def addOperators(num, target):
2    def backtrack(index, prev_operand, current_value, expression):
3        if index == len(num):
4            if current_value == target:
5                results.append(expression)
6            return
7        for i in range(index + 1, len(num) + 1):
8            current_str = num[index:i]
9            if len(current_str) > 1 and current_str[0] == '0':
10                continue
11            current_num = int(current_str)
12            backtrack(i, current_num, current_value + current_num, expression + '+' + current_str)
13            backtrack(i, -current_num, current_value - current_num, expression + '-' + current_str)
14            backtrack(i, prev_operand * current_num, current_value - prev_operand + (prev_operand * current_num), expression + '*' + current_str)
15
16    results = []
17    for i in range(1, len(num) + 1):
18        backtrack(i, int(num[:i]), int(num[:i]), num[:i])
19    return results

ℹ

Complexity note: The time complexity is O(4^n) due to the branching factor of 4 (3 operators + 1 for no operator) at each digit position. The space complexity is O(n) for the recursion stack.

1Understanding operator precedence is crucial when building expressions.
2Backtracking is a powerful technique for exploring combinations and permutations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.