## Microsoft Interview Question

SDE-3s**Country:**United States

**Interview Type:**In-Person

Copy pasting a comment of fun4LeetCode from leetcode.

General structure

```
public int calculate(String s) {
int l1 = 0, o1 = 1; // Initialization of level one
int l2 = 1, o2 = 1; // Initialization of level two
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c is a digit) {
--> we have an operand of type number, so find its value "num"
--> then evaluate at level two: l2 = (o2 == 1 ? l2 * num : l2 / num);
} else if (c is a lowercase letter) {
--> we have an operand of type variable, so find its name "var"
--> then look up the variable mapping table to find its value "num";
--> lastly evaluate at level two: l2 = (o2 == 1 ? l2 * num : l2 / num);
} else if (c is an opening parenthesis) {
--> we have an operand of type subexpression, so find its string representation
--> then recursively call the "calculate" function to find its value "num";
--> lastly evaluate at level two: l2 = (o2 == 1 ? l2 * num : l2 / num);
} else if (c is a level two operator) {
--> o2 needs to be updated: o2 = (c == '*' ? 1 : -1);
} else if (c is a level one operator) {
--> demotion happens here: l1 = l1 + o1 * l2;
--> o1 needs to be updated: o1 = (c == '+' ? 1 : -1);
--> l2, o2 need to be reset: l2 = 1, o2 = 1;
}
return (l1 + o1 * l2); // end of expression reached, so demotion happens again
}
```

Recursive solution: O(n^2) time, O(n) space

```
public int basicCalculatorIII(String s) {
int l1 = 0, o1 = 1;
int l2 = 1, o2 = 1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
int num = c - '0';
while (i + 1 < s.length() && Character.isDigit(s.charAt(i + 1))) {
num = num * 10 + (s.charAt(++i) - '0');
}
l2 = (o2 == 1 ? l2 * num : l2 / num);
} else if (c == '(') {
int j = i;
for (int cnt = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') cnt++;
if (s.charAt(i) == ')') cnt--;
if (cnt == 0) break;
}
int num = basicCalculatorIII(s.substring(j + 1, i));
l2 = (o2 == 1 ? l2 * num : l2 / num);
} else if (c == '*' || c == '/') {
o2 = (c == '*' ? 1 : -1);
} else if (c == '+' || c == '-') {
l1 = l1 + o1 * l2;
o1 = (c == '+' ? 1 : -1);
l2 = 1; o2 = 1;
}
}
return (l1 + o1 * l2);
}
```

1. Convert the infix expression to prefix

- Guru April 07, 20182. Evaluate prefix expression using stack.