227. Basic Calculator II LeetCode Solution

In this guide, you will get 227. Basic Calculator II LeetCode Solution with the best time and space complexity. The solution to Basic Calculator II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Basic Calculator II solution in C++
4. Basic Calculator II solution in Java
5. Basic Calculator II solution in Python
6. Additional Resources

Problem Statement of Basic Calculator II

Given a string s which represents an expression, evaluate this expression and return its value.
The integer division should truncate toward zero.
You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 – 1].
Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:
Input: s = “3+2*2″
Output: 7
Example 2:
Input: s = ” 3/2 ”
Output: 1
Example 3:
Input: s = ” 3+5 / 2 ”
Output: 5

Constraints:

1 <= s.length <= 3 * 105
s consists of integers and operators ('+', '-', '*', '/') separated by some number of spaces.
s represents a valid expression.
All the integers in the expression are non-negative integers in the range [0, 231 – 1].
The answer is guaranteed to fit in a 32-bit integer.

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

227. Basic Calculator II LeetCode Solution in C++

class Solution {
 public:
  int calculate(string s) {
    stack<int> nums;
    stack<char> ops;

    for (int i = 0; i < s.length(); ++i) {
      const char c = s[i];
      if (isdigit(c)) {
        int num = c - '0';
        while (i + 1 < s.length() && isdigit(s[i + 1])) {
          num = num * 10 + (s[i + 1] - '0');
          ++i;
        }
        nums.push(num);
      } else if (c == '+' || c == '-' || c == '*' || c == '/') {
        while (!ops.empty() && compare(ops.top(), c))
          nums.push(calculate(pop(ops), pop(nums), pop(nums)));
        ops.push(c);
      }
    }

    while (!ops.empty())
      nums.push(calculate(pop(ops), pop(nums), pop(nums)));

    return nums.top();
  }

 private:
  int calculate(char op, int b, int a) {
    switch (op) {
      case '+':
        return a + b;
      case '-':
        return a - b;
      case '*':
        return a * b;
      case '/':
        return a / b;
    }
    throw;
  }

  // Returns true if priority(op1) >= priority(op2).
  bool compare(char op1, char op2) {
    return op1 == '*' || op1 == '/' || op2 == '+' || op2 == '-';
  }

  char pop(stack<char>& ops) {
    const char op = ops.top();
    ops.pop();
    return op;
  }

  int pop(stack<int>& nums) {
    const int num = nums.top();
    nums.pop();
    return num;
  }
};
/* code provided by PROGIEZ */

227. Basic Calculator II LeetCode Solution in Java

class Solution {
  public int calculate(String s) {
    Deque<Integer> nums = new ArrayDeque<>();
    Deque<Character> ops = new ArrayDeque<>();

    for (int i = 0; i < s.length(); ++i) {
      final 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 + 1) - '0');
          ++i;
        }
        nums.push(num);
      } else if (c == '+' || c == '-' || c == '*' || c == '/') {
        while (!ops.isEmpty() && compare(ops.peek(), c))
          nums.push(calculate(ops.pop(), nums.pop(), nums.pop()));
        ops.push(c);
      }
    }

    while (!ops.isEmpty())
      nums.push(calculate(ops.pop(), nums.pop(), nums.pop()));

    return nums.peek();
  }

  private int calculate(char op, int b, int a) {
    switch (op) {
      case '+':
        return a + b;
      case '-':
        return a - b;
      case '*':
        return a * b;
      case '/':
        return a / b;
    }
    throw new IllegalArgumentException();
  }

  // Returns true if priority(op1) >= priority(op2).
  private boolean compare(char op1, char op2) {
    return op1 == '*' || op1 == '/' || op2 == '+' || op2 == '-';
  }
}
// code provided by PROGIEZ

227. Basic Calculator II LeetCode Solution in Python

class Solution:
  def calculate(self, s: str) -> int:
    ans = 0
    prevNum = 0
    currNum = 0
    op = '+'

    for i, c in enumerate(s):
      if c.isdigit():
        currNum = currNum * 10 + int(c)
      if not c.isdigit() and c != ' ' or i == len(s) - 1:
        if op == '+' or op == '-':
          ans += prevNum
          prevNum = currNum if op == '+' else -currNum
        elif op == '*':
          prevNum = prevNum * currNum
        elif op == '/':
          if prevNum < 0:
            prevNum = math.ceil(prevNum / currNum)
          else:
            prevNum = prevNum // currNum
        op = c
        currNum = 0

    return ans + prevNum
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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