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

class Solution {
    public static void minMaxValues(String exp) {
        // Converting String into array of Strings
        ArrayList<StringBuilder> list = new ArrayList<>();
        for (int i = 0; i < exp.length(); i++) {
            StringBuilder str = new StringBuilder();
            if (exp.charAt(i) == '+' || exp.charAt(i) == '*')
                str.append(exp.charAt(i));
            else {
                while (i < exp.length() && exp.charAt(i) >= '0' && exp.charAt(i) <= '9')
                    str.append(exp.charAt(i++));
                i--;
            }
            list.add(str);
        }
        String[] str = new String[list.size()];
        for (int i = 0; i < list.size(); i++)
            str[i] = list.get(i).toString();
        int n = str.length;
        // DP[i][j] -> Min/Max ans between [i,j]
        int[][] minDP = new int[n][n], maxDP = new int[n][n];
        // Diagonal base case
        for (int i = 0; i < n; i += 2)
            minDP[i][i] = maxDP[i][i] = Integer.parseInt(str[i]);
        // Remember i,j stays at numerical values
        for (int i = n - 1; i >= 0; i -= 2) {
            for (int j = i + 2; j < n; j += 2) {
                minDP[i][j] = Integer.MAX_VALUE;
                maxDP[i][j] = Integer.MIN_VALUE;
                // Consider all the symbols one be one for first execution
                // K runs at symbol values
                for (int k = i + 1; k <= j - 1; k += 2) {
                    int leftAnsMin = minDP[i][k - 1], rightAnsMin = minDP[k + 1][j];
                    int leftAnsMax = maxDP[i][k - 1], rightAnsMax = maxDP[k + 1][j];
                    int minAns, maxAns;
                    if (str[k].equals("+")) {
                        minAns = leftAnsMin + rightAnsMin;
                        maxAns = leftAnsMax + rightAnsMax;
                    } else {
                        minAns = leftAnsMin * rightAnsMin;
                        maxAns = leftAnsMax * rightAnsMax;
                    }
                    minDP[i][j] = Math.min(minDP[i][j], minAns);
                    maxDP[i][j] = Math.max(maxDP[i][j], maxAns);
                }
            }
        }
        System.out.println("Max Result : " + maxDP[0][n - 1]);
        System.out.println("Min Result : " + minDP[0][n - 1]);
    }
}