WEEK#13 Divide and Conquer_Different Ways to Add Parentheses

Description of the Problem

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1
Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2
Output: [0, 2]

Example 2
Input: "2x3-4x5"

(2x(3-(4x5))) = -34
((2x3)-(4x5)) = -14
((2x(3-4))x5) = -10
(2x((3-4)x5)) = -10
(((2x3)-4)x5) = 10
Output: [-34, -14, -10, -10, 10]

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Solution1 : Divide and Conquer (Wrong Answer 19/25)

Decompose the problem into smaller and similar parts.
Realizing that the result is certain when there are 2 operators, so we want to make whatever input we get into the pattern of 2 operators.
For example, an expression of 4 operators A op B op C op D can be divided into expressions of 3 operators:

1. (A op B) op C op D
2. A op (B op C) op D
3. A op B op (C op D)

Which can be furtherly divided into expressions of 2 operators:
Take 1 for example:
1.1 ((A op B) op C) op D
1.2 (A op B) op (C op D)

Whose result would then be unique.

class Solution {
public:
    vector diffWaysToCompute(string input) {

        vector results;
        vector operators;
        vector operations;
        vector expressions;

        bool flag; // check if the input has only one operator.
        for (int i = 0; i < input.size(); i++) {
            if (input[i] == '+' || input[i] == '-' || input[i] == '*') {
                flag = false;
                break;
            }
            flag = true;
        }
        if (flag) {
            results.push_back(stoi(input));
            return results;
        }
        int LastBegin = 0; // begin index of the last operator
        string oprt = "";
        for (int i = 0; i < input.length(); i++) { // get all operators and operations.
            if (input[i] == '+' || input[i] == '-' || input[i] == '*') {
                operations.push_back(input[i]);
                oprt = input.substr(LastBegin, i-LastBegin);
                operators.push_back(stoi(oprt));
                LastBegin = i + 1;
            }
        }
        oprt = input.substr(LastBegin, input.length() - LastBegin);
        operators.push_back(stoi(oprt));
        DivideAndConquer(operators, operations, results, expressions);

        return results;
    }

    void DivideAndConquer(vector operators, vector operations, vector& results, vector& expressions) {
        int result;
        if (operators.size() == 2) { // unique result
            stringstream ss;
            string expression = "";
            if (operations[0] == '+') {
                result = operators[0] + operators[1];
                ss << (operators[0]);
                ss << '+';
                ss << (operators[1]);
                expression = ss.str();
            }
            else if (operations[0] == '-') {
                result = operators[0] - operators[1];
                ss<<(operators[0]);
                ss<< '-';
                ss<<(operators[1]);
                expression = ss.str();
            }
            else if (operations[0] == '*') {
                result = operators[0] * operators[1];
                ss << (operators[0]);
                ss << '*';
                ss << (operators[1]);
                expression = ss.str();
            }

            if (!FindExpression(expression, expressions)) {
                results.push_back(result);
                expressions.push_back(expression);
            }
            return;
        }


        vector NextRoundOperators;
        vector NextRoundOperations;

        for (int i = 0; i < operations.size(); i++) { // calculate in different orders
            NextRoundOperations.clear();
            NextRoundOperators.clear();
            if (operations[i] == '+')
                result = operators[i] + operators[i + 1];
            else if (operations[i] == '-')
                result = operators[i] - operators[i + 1];
            else if (operations[i] == '*')
                result = operators[i] * operators[i + 1];
            for (int j = 0; j< operators.size(); j++) {
                if (j != i && j != i + 1)
                    NextRoundOperators.push_back(operators[j]);
            }
            vector::iterator it = NextRoundOperators.begin();
            int temp = i;
            while (temp--)
                it++;
            NextRoundOperators.insert(it, result);
            for (int k = 0; k < operations.size(); k++) {
                if (k != i)
                    NextRoundOperations.push_back(operations[k]);
            }
            DivideAndConquer(NextRoundOperators, NextRoundOperations, results, expressions);
        }

    }

    bool FindExpression(string exp, vector exps) {
        for (int i = 0; i < exps.size(); i++) {
            if (exp == exps[i])
                return true;
        }
        return false;
    }

};

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Solution 2

Every time we encounter an operator, record it and divide the expression into 2 parts [exp1 op exp2] by the operator.
For each part, which is also an expression, would have its calculating result, so the expression [exp1 op exp2] would also have its result.

class Solution {
public:
    vector diffWaysToCompute(string input) {
        vector result;
        int size = input.size();
        for (int i = 0; i < size; i++) {
            char cur = input[i];
            if (cur == '+' || cur == '-' || cur == '*') {
                // Split input string into two parts and solve them recursively
                vector result1 = diffWaysToCompute(input.substr(0, i));
                vector result2 = diffWaysToCompute(input.substr(i+1));
                for (auto n1 : result1) {
                    for (auto n2 : result2) {
                        if (cur == '+')
                            result.push_back(n1 + n2);
                        else if (cur == '-')
                            result.push_back(n1 - n2);
                        else
                            result.push_back(n1 * n2);    
                    }
                }
            }
        }
        // if the input string contains only number
        if (result.empty())
            result.push_back(atoi(input.c_str()));
        return result;
    }
};