# C++ 4ms DP solution

• The recursive solution is not so efficient since it computes the possible results of the same substrings many times. To improve the time complexity, we can use dp which stores the intermediate results of the substrings. Define a two-dimensional array

dp[i][j] = vector of all possible results computed from the substring starting from operand i and
including the subsequent j operators and operands.

Let cntOperators be the count of operators in the input string. Note that the count of operands is cntOperators + 1. The range of i is [0, cntOperators + 1) and the range of j is [0, cntOperators).

``````class Solution
{
// Separate the operands and the operators ('+', '-', '*')
// in the input string.
void GetOperandsAndOperatorsFromString(
string &input,
vector<int> &operands,
vector<char> &operators)
{
int number = 0;
for (int i = 0; i <= input.length(); i++)
{
if ((input[i] >= '0') && (input[i] <= '9'))
{
// This character is a digit and add it to
// the number.
number = 10*number + (input[i] - '0');
}
else if ((input[i] == '+') ||
(input[i] == '-') ||
(input[i] == '*'))
{
// This character is an operator and it implies
// the end of the previous number. Add the number
// to the operand vector and reset the number. Also
// add the operator to the operator vector.
operands.push_back(number);
number = 0;

operators.push_back(input[i]);
}
else if (input[i] == '\0')
{
// This is the end of the input string. Add the
// last number to the operand vector.
operands.push_back(number);
}
}
}

public:
vector<int> diffWaysToCompute(string input)
{
vector<int> operands;
vector<char> operators;

// Get the operands and operators from the input string.
GetOperandsAndOperatorsFromString(input, operands, operators);

// Note that count of operands = cnt of operators + 1.
int cntOperators = operators.size();

// dp[operandIndex][cntUsedOperators] = vector of all possible results
// computed from the substring starting from operand "operandIndex" and
// including the subsequent "cntUsedOperators" operators and operands.
vector<vector<vector<int>>> dp(cntOperators + 1, vector<vector<int>>());

// When cntUsedOperators = 0, dp[operandIndex][0] = operands[operandIndex].
for (int operandIndex = 0; operandIndex < cntOperators + 1; operandIndex++)
{
dp[operandIndex].push_back(vector<int>{operands[operandIndex]});
}

// cntUsedOperators = 1,...,cntOperators.
for (int cntUsedOperators = 1; cntUsedOperators <= cntOperators; cntUsedOperators++)
{
// Note that operandIndex starts from 0 and ends at the last operand which
// is followed by "cntUsedOperators" operators and operands.
for (int operandIndex = 0;
operandIndex < cntOperators + 1 - cntUsedOperators;
operandIndex++)
{
// dp[operandIndex][cntUsedOperators] = the union of all outcomes for each
// possible last operator.
vector<int> tmpRes;
for (int lastOperatorIndex = operandIndex; lastOperatorIndex < operandIndex + cntUsedOperators; lastOperatorIndex++)
{
// The last operator breaks the substring into two halves.
// dp[operandIndex][lastOperatorIndex - operandIndex] =
//  all the outcomes of the first half.
// dp[lastOperatorIndex + 1][cntUsedOperators - 1 - (lastOperatorIndex - operandIndex)] =
//  all the outcomes of the second half.
for (auto &x : dp[operandIndex][lastOperatorIndex - operandIndex])
{
for (auto &y : dp[lastOperatorIndex + 1][cntUsedOperators - 1 - (lastOperatorIndex - operandIndex)])
{
if (operators[lastOperatorIndex] == '+')
{
tmpRes.push_back(x + y);
}
else if (operators[lastOperatorIndex] == '-')
{
tmpRes.push_back(x - y);
}
else
{
tmpRes.push_back(x * y);
}
}
}
}

dp[operandIndex].push_back(tmpRes);
}
}

// The final result is all possible outcomes starting from operand 0 and including all operators.
return dp[0][cntOperators];
}
};``````

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