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Different Ways to Add Parentheses

Updated on 23 May, 2025
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Problem Statement

In the given problem, we are provided with a string named expression, which contains integers and the operators '+', '-', and '*'. The core task is to calculate every possible result one can get by applying all the different feasible groupings of numbers and operators in the expression. These results can be in any sequence. This challenge not only tests basic computation but also how these numbers and operators can be grouped in various ways to yield different outcomes. The constraints ensure that the outcomes are manageable in size, as they fit within a 32-bit integer, and emphasize that the unique results will not be excessively numerous, capped at 10,000 different outcomes.

Examples

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.

Approach and Intuition

This problem is a classic example of using recursion to handle different parts of the problem based on splitting the expression around an operator and then recursively solving each of those parts. Here's how we can approach it:

Step-by-step Process

  1. Base Case Identification:

    • If the string does not contain any operator, it means the string is a single number. In this case, we simply return the number itself as the only possible result.

  2. Recursive Breakdown:

    • For each operator in the string:
      • Divide the expression into two parts, left and right, from the current operator.
      • Recursively solve for all results from the left part and the right part.

  3. Combining Results:

    • After obtaining all possible results from the two halves, combine these using the current operator. For instance, if you are at operator '+', compute the sum for each paired result from the left and right lists.
    • Store these combined results.

  4. Memoization to Optimize:

    • Since expressions might be recomputed, especially in cases with recurring subexpressions, store previously computed results for specific subexpressions to avoid redundant calculations.

Understanding with Examples

This approach works conveniently with the provided examples:

  • Example 1 (expression = "2-1-1"):

    • Break down around the first '-' to get "2" and "1-1".
    • "2" alone results in [2].
    • "1-1" recursively breaks to yield [0].
    • Compute (2-0) and (2-(1-1)) to get results [2, 0].

  • Example 2 (expression = "2*3-4*5"):

    • This longer expression can be recursively broken down around each operator to evaluate expressions like "23", "3-4", and "45" individually and then combined.
    • The resulting different computed values from these combinations showcase the diverse results stemming from different grouping possibilities offered by recursive breakdowns and calculations.

Constraints Consideration

  • The length constraint (1 to 20) makes recursive depth manageable, as the expression is not excessively long.
  • Given that the only allowed operators are '+', '-', and '*', parsing and computation revolve only around basic arithmetic, making a straightforward recursive approach feasible without dealing with operator precedence complications beyond these three.
  • The number range (0 to 99) for any single number ensures that individual results are within a 32-bit integer limit, aligning with typical calculation constraints in many programming scenarios.

Solutions

  • C++
  • Java
  • Python
cpp 
class Solution {
public:
    vector<int> computeExpression(string input) {
        int length = input.size();
        vector<vector<vector<int>>> cache(length, vector<vector<int>>(length));

        setupInitialConditions(input, cache);

        for (int len = 3; len <= length; len++) {
            for (int start = 0; start + len - 1 < length; start++) {
                int end = start + len - 1;
                evaluateSubexpression(input, cache, start, end);
            }
        }

        return cache[0][length - 1];
    }

private:
    void setupInitialConditions(string& input, vector<vector<vector<int>>>& cache) {
        int len = input.length();
        for (int i = 0; i < len; i++) {
            if (isdigit(input[i])) {
                int singleDigit = input[i] - '0';
                if (i + 1 < len && isdigit(input[i + 1])) {
                    int nextDigit = input[i + 1] - '0';
                    int twoDigitNum = singleDigit * 10 + nextDigit;
                    cache[i][i + 1].push_back(twoDigitNum);
                }
                cache[i][i].push_back(singleDigit);
            }
        }
    }

    void evaluateSubexpression(string& input, vector<vector<vector<int>>>& cache, int start, int end) {
        for (int divide = start; divide <= end; divide++) {
            if (isdigit(input[divide])) continue;
            vector<int> leftPartResults = cache[start][divide - 1];
            vector<int> rightPartResults = cache[divide + 1][end];

            calculateResults(input[divide], leftPartResults, rightPartResults, cache[start][end]);
        }
    }

    void calculateResults(char operation, vector<int>& leftValues, vector<int>& rightValues, vector<int>& accumulatedResults) {
        for (int lVal : leftValues) {
            for (int rVal : rightValues) {
                if (operation == '+')
                    accumulatedResults.push_back(lVal + rVal);
                else if (operation == '-')
                    accumulatedResults.push_back(lVal - rVal);
                else if (operation == '*')
                    accumulatedResults.push_back(lVal * rVal);
            }
        }
    }
};

The provided C++ code defines a solution for evaluating different possible outcomes of arithmetic expressions with varying placements of parentheses. The class Solution contains methods that deal with parsing, storing, and computing these outcomes based on combined recursive solutions and dynamic programming.

  • The core member function computeExpression initializes a 3-dimensional vector cache to store intermediate results and systematically increases the length of subtended expressions being examined.

  • setupInitialConditions parses the input string and populates cache entries for initial conditions based on single and consecutive digit numbers separated by non-digit characters. It prepares base cases for dynamic programming.

  • The evaluateSubexpression function iteratively attempts to split the expression at every operator within a given range to recursively determine out all possible values for subexpressions formed by adding parentheses around operators.

  • Each combination of results from the left and right partitions of the expression is processed depending on the current operator using calculateResults. This function computes results based on the operator and accumulates these for each split position, thereby expanding the cache with possible results of expressions.

  • The approach primarily decomposes the problem into smaller subproblems and combines their solutions to achieve the result for a bigger problem, leveraging dynamic programming techniques to reduce redundant computation.

  • Overall, the implementation efficiently computes all the possible results of the given arithmetic expression allowing for all combinations of parentheses placements, by iteratively and recursively breaking down expressions and caching intermediate results.

java 
class ExpressionCalculator {

    public List<Integer> computeValues(String expr) {
        int len = expr.length();
        List<Integer>[][] dp = new ArrayList[len][len];

        initializeDpValues(expr, dp);

        for (int segment = 3; segment <= len; segment++) {
            for (int i = 0; i + segment - 1 < len; i++) {
                int j = i + segment - 1;
                evaluateSubexpressions(expr, dp, i, j);
            }
        }

        return dp[0][len - 1];
    }

    private void initializeDpValues(String expr, List<Integer>[][] dp) {
        int len = expr.length();
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len; j++) {
                dp[i][j] = new ArrayList<>();
            }
        }

        for (int i = 0; i < len; i++) {
            if (Character.isDigit(expr.charAt(i))) {
                int num = expr.charAt(i) - '0';
                if (i + 1 < len && Character.isDigit(expr.charAt(i + 1))) {
                    int nextNum = expr.charAt(i + 1) - '0';
                    int combinedNum = 10 * num + nextNum;
                    dp[i][i + 1].add(combinedNum);
                }
                dp[i][i].add(num);
            }
        }
    }

    private void evaluateSubexpressions(String expr, List<Integer>[][] dp, int start, int end) {
        for (int k = start; k < end; k++) {
            if (!Character.isDigit(expr.charAt(k))) {
                List<Integer> leftParts = dp[start][k - 1];
                List<Integer> rightParts = dp[k + 1][end];

                calculateCombinations(expr.charAt(k), leftParts, rightParts, dp[start][end]);
            }
        }
    }

    private void calculateCombinations(char operator, List<Integer> leftVals, List<Integer> rightVals, List<Integer> outcome) {
        for (int left : leftVals) {
            for (int right : rightVals) {
                switch (operator) {
                    case '+':
                        outcome.add(left + right);
                        break;
                    case '-':
                        outcome.add(left - right);
                        break;
                    case '*':
                        outcome.add(left * right);
                        break;
                }
            }
        }
    }
}

Here's how the solution for the "Different Ways to Add Parentheses" is implemented in Java:

  • The code uses dynamic programming to compute different possible results from expressions by adding parentheses in various positions. It manages the computed values using a 2D list called dp.

  • Begin by initializing all cells in dp using the initializeDpValues method. This method assigns a list to every dp[i][j]. It also directly updates the dp array for simple single or double-digit numbers found in the expression.

  • The main computation occurs in computeValues method:

    1. Iterate over possible lengths of subexpressions.
    2. For each subexpression defined by its start and end indices (i and j), call evaluateSubexpressions to compute results based on different ways of adding parentheses.

  • Inside evaluateSubexpressions, iterate through each position in the subexpression:

    1. If the character at position k is an operator (+, -, *), then use the already computed results of the subexpressions to the left and right of the operator.
    2. Call calculateCombinations to compute all possible results for the current subexpression by applying the operator between every pair of results from the left and right subexpressions and store these results back in dp[start][end].

  • The calculateCombinations method handles the actual arithmetic operations. It iterates over all possible pairs of integers from left and right subexpressions and applies the current operator.

The last step in computeValues returns the list of results from dp[0][len-1], which contains all the possible outcomes for the complete expression set by varying the placement of parentheses. This approach ensures that all combinations are considered and the problem is solved in a structured manner using dynamic programming principles.

python 
class Calculator:
    def computeWays(self, input_str: str) -> List[int]:
        length = len(input_str)
        # Initialize dp storage
        memo = [[[] for _ in range(length)] for _ in range(length)]

        self._set_base_cases(input_str, memo)

        # Build the memo table for subproblems
        for size in range(3, length + 1):
            for start_idx in range(length - size + 1):
                end_idx = start_idx + size - 1
                self._compute_subproblems(input_str, memo, start_idx, end_idx)

        # Pull results for the full expression
        return memo[0][length - 1]

    def _set_base_cases(self, input_str: str, memo: List[List[List[int]]]):
        # Initialize values for single and two-digit numbers
        for idx, char in enumerate(input_str):
            if char.isdigit():
                digit = int(char)
                # Handling two-digit numbers
                if idx + 1 < len(input_str) and input_str[idx + 1].isdigit():
                    next_digit = int(input_str[idx + 1])
                    num = digit * 10 + next_digit
                    memo[idx][idx + 1].append(num)
                # Single-digit number
                memo[idx][idx].append(digit)

    def _compute_subproblems(self, input_str: str, memo: List[List[List[int]]], start: int, end: int):
        # Evaluate subexpressions splitting at each operator
        for split_point in range(start, end + 1):
            if input_str[split_point].isdigit():
                continue

            left_evaluations = memo[start][split_point - 1]
            right_evaluations = memo[split_point + 1][end]

            self._evaluate_operation(input_str[split_point], left_evaluations, right_evaluations, memo[start][end])

    def _evaluate_operation(self, operator: str, left_evaluations: List[int], right_evaluations: List[int], result_list: List[int]):
        # Evaluate result based on the operator and add to the result list
        for left in left_evaluations:
            for right in right_evaluations:
                if operator == '+':
                    result_list.append(left + right)
                elif operator == '-':
                    result_list.append(left - right)
                elif operator == '*':
                    result_list.append(left * right)

Explore different ways to add parentheses in an arithmetic expression to achieve different outcomes using Python. The provided Python script, structured within a Calculator class, implements a method computeWays to solve this problem. The approach involves dynamic programming to efficiently evaluate the expression for all possible results, given the operations: addition, subtraction, and multiplication.

Features of the Script:

  • The computeWays function receives an arithmetic expression as a string and initializes a dynamic programming table (memo) to store results of subexpressions.
  • Base cases for single and two-digit numbers are set using the _set_base_cases method. This method also ensures that consecutive digits are treated as multi-digit numbers.
  • The _compute_subproblems processes each possible subexpression within given limits, recursively determining all possible results from splitting the expression at each operation.

Step-by-step Operations:

  1. Initialize a memoization table to store intermediate results of subexpressions. Each cell contains a list to hold multiple possible evaluation results.
  2. Base cases are established for single and double-digit values.
  3. Loop through each subexpression of increasing sizes, solving for all possible results by splitting at each operator and combining the results based on the current operator.
  4. Utilize the _evaluate_operation for computing results based on current operation (either '+', '-', or '*') and aggregate them into the memoization table.

This method ensures that all possible combinations of adding parentheses are considered without recalculating the results of the same subexpressions repeatedly, leveraging memoization for optimization.

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