# Minimum and Maximum values of an expression with * and +

Given an expression which contains numbers and two operators ‘+’ and ‘*’, we need to find maximum and minimum value which can be obtained by evaluating this expression by different parenthesization.
Examples:

```Input  : expr = “1+2*3+4*5”
Output : Minimum Value = 27, Maximum Value = 105
Explanation:
Minimum evaluated value = 1 + (2*3) + (4*5) = 27
Maximum evaluated value = (1 + 2)*(3 + 4)*5 = 105
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can solve this problem by dynamic programming method, we can see that this problem is similar to matrix chain multiplication, here we are trying different parenthesization to maximize and minimize expression value instead of number of matrix multiplication.
In below code first we have separated the operators and numbers from given expression then two 2D arrays are taken for storing the intermediate result which are updated similar to matrix chain multiplication and different parenthesization are tried among the numbers but according to operators occurring in between them. At the end last cell of first row will store the final result in both the 2D arrays.

 `// C++ program to get maximum and minimum ` `// values of an expression ` `#include ` `using` `namespace` `std; ` ` `  `// Utility method to check whether a character ` `// is operator or not ` `bool` `isOperator(``char` `op) ` `{ ` `    ``return` `(op == ``'+'` `|| op == ``'*'``); ` `} ` ` `  `// method prints minimum and maximum value ` `// obtainable from an expression ` `void` `printMinAndMaxValueOfExp(string ``exp``) ` `{ ` `    ``vector<``int``> num; ` `    ``vector<``char``> opr; ` `    ``string tmp = ``""``; ` ` `  `    ``//  store operator and numbers in different vectors ` `    ``for` `(``int` `i = 0; i < ``exp``.length(); i++) ` `    ``{ ` `        ``if` `(isOperator(``exp``[i])) ` `        ``{ ` `            ``opr.push_back(``exp``[i]); ` `            ``num.push_back(``atoi``(tmp.c_str())); ` `            ``tmp = ``""``; ` `        ``} ` `        ``else` `        ``{ ` `            ``tmp += ``exp``[i]; ` `        ``} ` `    ``} ` `    ``//  storing last number in vector ` `    ``num.push_back(``atoi``(tmp.c_str())); ` ` `  `    ``int` `len = num.size(); ` `    ``int` `minVal[len][len]; ` `    ``int` `maxVal[len][len]; ` ` `  `    ``//  initializing minval and maxval 2D array ` `    ``for` `(``int` `i = 0; i < len; i++) ` `    ``{ ` `        ``for` `(``int` `j = 0; j < len; j++) ` `        ``{ ` `            ``minVal[i][j] = INT_MAX; ` `            ``maxVal[i][j] = 0; ` ` `  `            ``//  initializing main diagonal by num values ` `            ``if` `(i == j) ` `                ``minVal[i][j] = maxVal[i][j] = num[i]; ` `        ``} ` `    ``} ` ` `  `    ``// looping similar to matrix chain multiplication ` `    ``// and updating both 2D arrays ` `    ``for` `(``int` `L = 2; L <= len; L++) ` `    ``{ ` `        ``for` `(``int` `i = 0; i < len - L + 1; i++) ` `        ``{ ` `            ``int` `j = i + L - 1; ` `            ``for` `(``int` `k = i; k < j; k++) ` `            ``{ ` `                ``int` `minTmp = 0, maxTmp = 0; ` ` `  `                ``// if current operator is '+', updating tmp ` `                ``// variable by addition ` `                ``if``(opr[k] == ``'+'``) ` `                ``{ ` `                    ``minTmp = minVal[i][k] + minVal[k + 1][j]; ` `                    ``maxTmp = maxVal[i][k] + maxVal[k + 1][j]; ` `                ``} ` ` `  `                ``// if current operator is '*', updating tmp ` `                ``// variable by multiplication ` `                ``else` `if``(opr[k] == ``'*'``) ` `                ``{ ` `                    ``minTmp = minVal[i][k] * minVal[k + 1][j]; ` `                    ``maxTmp = maxVal[i][k] * maxVal[k + 1][j]; ` `                ``} ` ` `  `                ``//  updating array values by tmp variables ` `                ``if` `(minTmp < minVal[i][j]) ` `                    ``minVal[i][j] = minTmp; ` `                ``if` `(maxTmp > maxVal[i][j]) ` `                    ``maxVal[i][j] = maxTmp; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``//  last element of first row will store the result ` `    ``cout << ``"Minimum value : "` `<< minVal[len - 1] ` `         ``<< ``", Maximum value : "` `<< maxVal[len - 1]; ` `} ` ` `  `//  Driver code to test above methods ` `int` `main() ` `{ ` `    ``string expression = ``"1+2*3+4*5"``; ` `    ``printMinAndMaxValueOfExp(expression); ` `    ``return` `0; ` `} `

Output:

```Minimum value : 27, Maximum value : 105
```

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Improved By : Rajat Singh 11