Expression Add Operators

Given a string num that contains only digits and an integer target, find all possible ways to add the binary operators '+', '-', or '*' between the digits of num so that the resultant expression evaluates to the target value.

You should return a list of strings; each string should be a valid expression.

For example:

• Example 1:

  • num = "123", target = 6
  • Output: ["123", "1+2+3"]

• Example 2:

  • num = "232", target = 8
  • Output: ["23+2", "2+32"]

• Example 3:

  • num = "105", target = 5
  • Output: ["1*0+5","10-5"]

• Example 4:

  • num = "00", target = 0
  • Output: ["0*0", "0+0", "0-0"]

• Example 5:

  • num = "3456237490", target = 9191
  • Output: []

Can you describe your approach to solving this problem, and then provide a code implementation?

Expression Add Operators

Let's explore how to solve the "Expression Add Operators" problem. This involves inserting binary operators (+, -, *) between the digits of a given string of numbers to evaluate to a target number.

1. Naive (Brute Force) Solution

The most straightforward approach is to try all possible combinations of operators between the digits. We can use recursion to explore all combinations. At each digit, we have three choices: add a +, a -, a *, or no operator (which means we concatenate the current digit with the previous number). We continue this process until we reach the end of the input string, and then we evaluate the expression to see if it matches the target.

This approach involves generating all possible expressions and then evaluating them.

2. Optimal Solution

The optimal solution still uses recursion and backtracking, but it's more efficient because it avoids redundant calculations and keeps track of intermediate results. The key optimization is to maintain the following state during the recursion:

• current_expression: The expression string being built.
• current_value: The value of the expression so far.
• previous_operand: The value of the last operand added (important for handling multiplication correctly).
• index: The current index in the input string.

When we encounter a multiplication, we need to update the current_value by subtracting the previous_operand and adding the result of previous_operand * current_operand. This ensures that multiplication has precedence.

3. Big(O) Run-time

The time complexity is difficult to precisely determine due to the nature of backtracking. In the worst case, for a string of length n, there are 3^(n-1) possible expressions (since there are three choices of operators or no operator between each digit). The expression evaluation contributes to the time complexity as well. So, the time complexity is approximately O(3^n).

4. Big(O) Space Usage

The space complexity is dominated by the recursion depth, which can be O(n) in the worst case where n is the length of the input string. We also store the current_expression, which can also be O(n). Therefore, the space complexity is O(n). Also, in Java, Strings are immutable, and so many different String objects can be created throughout the function call which can use up some memory.

5. Edge Cases

• Leading Zeros: We need to handle cases where the input number string contains leading zeros. We should avoid considering numbers with leading zeros (e.g., "05") as valid operands unless the operand is just "0".
• Overflow/Underflow: The intermediate results of calculations might overflow or underflow, so we should use long integers to avoid this issue, before comparing at the end with the target.
• Empty Input: Handle the case where the input string is empty.
• Single Digit Input: Handle cases where input string contains a single digit.
• Target equals zero: The target can equal zero, and this case must be handled correctly.

6. Code (Java)

java import java.util.ArrayList; import java.util.List;

class Solution { public List<String> addOperators(String num, int target) { List<String> result = new ArrayList<>(); if (num == null || num.length() == 0) { return result; } helper(result, "", num, target, 0, 0, 0); return result; }

private void helper(List<String> result, String currentExpression, String num, int target, int index, long currentValue, long previousOperand) {
    if (index == num.length()) {
        if (currentValue == target) {
            result.add(currentExpression);
        }
        return;
    }

    for (int i = index; i < num.length(); i++) {
        if (i > index && num.charAt(index) == '0') {
            break; // Skip leading zeros
        }
        long currentOperand = Long.parseLong(num.substring(index, i + 1));

        if (index == 0) {
            helper(result, String.valueOf(currentOperand), num, target, i + 1, currentOperand, currentOperand);
        } else {
            helper(result, currentExpression + "+" + currentOperand, num, target, i + 1, currentValue + currentOperand, currentOperand);
            helper(result, currentExpression + "-" + currentOperand, num, target, i + 1, currentValue - currentOperand, -currentOperand);
            helper(result, currentExpression + "*" + currentOperand, num, target, i + 1, currentValue - previousOperand + previousOperand * currentOperand, previousOperand * currentOperand);
        }
    }
}

}

This Java code implements the optimal solution using recursion and backtracking, considering edge cases like leading zeros and using long to prevent overflow. The helper function recursively builds the expression and explores different operator combinations.