Problem editorial

Basic Calculator III

HardStack PatternsMathStringStack

Editorial

Core insight

Basic Calculator III · Stack Patterns

Core Insight for Basic Calculator III

The key intuition for solving LeetCode 772 efficiently is to treat every expression (or sub-expression within parentheses) as a sequence of additions.

Reduction to Summation: We can treat subtraction A - B as A + (-B). This allows us to reduce the final step of any scope to simply summing up the stack.
Greedy Evaluation of High Precedence: When we encounter multiplication * or division /, we cannot wait. These operators bind tighter than + or -. We must pop the top value from our stack, perform the operation with the current number, and push the result back.
Delayed Evaluation of Low Precedence: When we encounter + or -, we push the previous number onto the stack and update our "pending operator." We do not evaluate the sum yet because a future * or / might need the current number.
Recursion for Parentheses: When we see a (, we are entering a sub-problem. This sub-problem is identical to the main problem but ends at ). We can recursively call our solver (utilizing the system call stack) to evaluate the sub-expression and return a single integer.

Visual Description: Imagine a stack representing the current "summation terms." If the expression is 2 + 3 * 4, the stack evolves as follows:

Read 2, operator is +. Push 2. Stack: [2].
Read 3, operator is *. Push 3. Stack: [2, 3].
Read 4. Previous operator was *. Pop 3, calculate 3 * 4 = 12. Push 12. Stack: [2, 12].
End of string. Sum the stack: 2 + 12 = 14.

The stack effectively isolates terms that are separated by + signs, while * and / merge terms immediately before they are finalized on the stack.

Deep dive

Full breakdown

Full reasoning, pitfalls, and implementation flow

LeetCode 772: Basic Calculator III Solution & Explanation

Problem Overview

TL;DR: The optimal solution uses recursion (implicit stack) to handle parentheses and an explicit stack to handle operator precedence, reducing the entire expression to a sum of integers.

The LeetCode 772 problem, Basic Calculator III, asks us to evaluate a mathematical expression string that includes non-negative integers, basic operators (+, -, *, /), and parentheses ( ). The evaluation must respect standard order of operations (multiplication and division before addition and subtraction) and handle nested parentheses. Integer division must truncate toward zero.

This is a comprehensive interview question because it combines the logic of Basic Calculator I (parentheses) and Basic Calculator II (precedence).

Brute Force Approach for Basic Calculator III

A naive or brute force approach might attempt to solve the expression by repeatedly scanning the string to find the operation with the highest precedence, executing it, and replacing that part of the string with the result.

Find the deepest set of parentheses.
Inside those parentheses, scan for * or /.
Perform the operation and create a new string with the result substituted.
Repeat until no * or / remain in that scope, then process + and -.
Remove the parentheses and repeat the process for the outer scope.

Pseudo-code:

text

while string has operators:
    find inner-most (...)
    solve multiplication/division inside
    solve addition/subtraction inside
    replace (...) with result in string
return result

Why it fails: This approach has a time complexity of O(N²) or worse because string concatenation and replacement are expensive operations ( $O(N)$ ) that are performed repeatedly. For a string length of $10^4$ , this will likely result in a Time Limit Exceeded (TLE) error. Furthermore, implementing the string parsing logic correctly without bugs is tedious and error-prone during an interview.

Algorithm Strategy: Stack Patterns

We will implement a recursive solution where each level of recursion solves the expression within a pair of parentheses (or the entire string).

State Maintenance:

Stack: Stores integers to be summed at the end of the current scope.
Current Number (num): Builds the multi-digit integer currently being parsed.
Last Operator (op): Tracks the operator that precedes the num. Initialized to +.

Constraints & Boundaries:

The recursion handles the boundaries of ( and ).
The loop terminates when we reach the end of the string or a closing parenthesis ).

Logic:

Iterate through the string (or a shared index/iterator).
If the character is a digit, update num.
If the character is (, recursively call the function to evaluate the sub-expression. The result becomes our num.
If the character is an operator +, -, *, / or a closing parenthesis ) or the end of the string:
- Apply op to the current num and the top of the stack.
  - +: Push num.
  - -: Push -num.
  - *: Pop top, push top * num.
  - /: Pop top, push top / num (truncate toward zero).
- Update op to the current character.
- Reset num to 0.
If the character is ), break the loop.
Return the sum of elements in the stack.

Execution Flow

Let's trace 3 + 2 * ( 5 - 1 )

Start Main Scope: op = '+', stack = [].
Parse 3. op is +. Push 3. stack = [3]. Update op = '+'.
Parse 2. op is +. Push 2. stack = [3, 2]. Update op = '*'.
Encounter (. Recurse.
- Sub-Scope: op = '+', stack = [].
- Parse 5. op is +. Push 5. stack = [5]. Update op = '-'.
- Parse 1. op is -. Push -1. stack = [5, -1]. Update op = ')'.
- Encounter ). Break loop.
- Return sum of stack: 5 + (-1) = 4.
Back to Main Scope: The recursion returned 4. So effectively, num = 4.
Previous op was *. Pop 2 from stack. Calculate 2 * 4 = 8. Push 8. stack = [3, 8].
End of string.
Return sum of stack: 3 + 8 = 11.

Proof of Correctness

The algorithm's correctness relies on the invariant that the stack always contains terms that are ready to be added.

Base Case: Single numbers are pushed directly (or negated for subtraction).
Inductive Step (Precedence): Multiplicative operators (*, /) combine the current number with the last pushed number immediately. This ensures standard order of operations because the result is placed back on the stack as a single term, effectively treating a + b * c as a + (b*c).
Inductive Step (Parentheses): Recursion treats a sub-expression (...) as a single atomic number. Since the sub-expression is evaluated using the same logic, it correctly returns the evaluated integer, maintaining the structural integrity of the expression.

Thus, at the end of any scope, the stack contains a list of values [v1, v2, ..., vn] representing v1 + v2 + ... + vn.

Pattern Reuse Notes

The Stack - Expression Evaluation pattern is highly reusable. Understanding the mechanism of "delaying evaluation" using a stack is key for these related problems:

LeetCode 150: Evaluate Reverse Polish Notation: This is the "backend" of the calculator. It evaluates expressions already in postfix format.
LeetCode 224: Basic Calculator: Handles +, -, and (). It is a subset of 772. The logic is identical, though strictly handling signs is sufficient without a full multiplication stack.
LeetCode 227: Basic Calculator II: Handles +, -, *, / but no parentheses. This is exactly the logic inside one "scope" of Basic Calculator III.

Feynman coach · Before you peek at the code

Can you explain the core insight in your own words?

If you can explain it before reading code, you’re far more likely to recall it under interview pressure.

Loading your attempt…

Reference Implementation

C++ Solution for LeetCode 772

cpp

1class Solution {
2public:
3    int calculate(string s) {
4        int i = 0;
5        return parse(s, i);
6    }
7
8private:
9    int parse(const string& s, int& i) {
10        vector<long> stack;
11        char op = '+';
12        long num = 0;
13
14        while (i < s.length()) {
15            char c = s[i++];
16            
17            if (isdigit(c)) {
18                num = num * 10 + (c - '0');
19            } else if (c == '(') {
20                num = parse(s, i);
21            }
22
23            // If c is an operator, or ')', or we are at the end, process the last number
24            // Note: We check i < s.length() inside the loop, but need to process
25            // the last number if we hit the end.
26            // A trick is to process if !isdigit(c) and c != ' ' or i == s.length()
27            // However, since we incremented i, we look at the previous char logic implicitly.
28            // Let's stick to the standard check:
29            if ((!isdigit(c) && c != ' ') || i == s.length()) { // Check current char (which is s[i-1] now due to i++)
30                 // Wait, simpler logic: check the char we just read
31                 // If the char we just read was '(', we already recursed and have num.
32                 // If it was ')', we process and break.
33            }
34            
35            // Correct logic flow for loop:
36            // The check needs to happen if c is an operator or end of block
37            // Since we incremented i, c is the character we are processing.
38            if ((!isdigit(c) && c != ' ') || i > s.length()) { 
39                // Wait, logic correction: standard loop structure is better.
40                // Let's rewind to the standard clean pattern.
41            }
42        }
43        return 0; 
44    }
45};
46
47// Re-writing strictly clean C++ implementation
48class Solution {
49public:
50    int calculate(string s) {
51        int i = 0;
52        return helper(s, i);
53    }
54
55    int helper(const string& s, int& i) {
56        vector<long> stack;
57        char op = '+';
58        long num = 0;
59
60        while (i < s.length()) {
61            char c = s[i++];
62
63            if (isdigit(c)) {
64                num = num * 10 + (c - '0');
65            } else if (c == '(') {
66                num = helper(s, i);
67            }
68
69            // Process if operator, closing parenthesis, or end of string
70            if (i == s.length() || c == '+' || c == '-' || c == '*' || c == '/' || c == ')') {
71                if (op == '+') {
72                    stack.push_back(num);
73                } else if (op == '-') {
74                    stack.push_back(-num);
75                } else if (op == '*') {
76                    long last = stack.back();
77                    stack.pop_back();
78                    stack.push_back(last * num);
79                } else if (op == '/') {
80                    long last = stack.back();
81                    stack.pop_back();
82                    // Integer division truncates toward zero in C++
83                    stack.push_back(last / num);
84                }
85                
86                op = c;
87                num = 0;
88
89                if (c == ')') break;
90            }
91        }
92
93        long result = 0;
94        for (long n : stack) result += n;
95        return (int)result;
96    }
97};

Java Solution for LeetCode 772

java

1import java.util.*;
2
3class Solution {
4    private int i = 0; // Global index to track position across recursion
5
6    public int calculate(String s) {
7        i = 0;
8        return helper(s);
9    }
10
11    private int helper(String s) {
12        Stack<Integer> stack = new Stack<>();
13        char op = '+';
14        int num = 0;
15
16        while (i < s.length()) {
17            char c = s.charAt(i++);
18
19            if (Character.isDigit(c)) {
20                num = num * 10 + (c - '0');
21            } else if (c == '(') {
22                num = helper(s);
23            }
24
25            // If operator, ')', or end of string (handled by loop condition + internal check)
26            // We check if the character is meaningful for processing logic
27            if (i >= s.length() || c == '+' || c == '-' || c == '*' || c == '/' || c == ')') {
28                if (op == '+') {
29                    stack.push(num);
30                } else if (op == '-') {
31                    stack.push(-num);
32                } else if (op == '*') {
33                    stack.push(stack.pop() * num);
34                } else if (op == '/') {
35                    stack.push(stack.pop() / num);
36                }
37                
38                op = c;
39                num = 0;
40
41                if (c == ')') break;
42            }
43        }
44
45        int result = 0;
46        for (int n : stack) {
47            result += n;
48        }
49        return result;
50    }
51}

Python Solution for LeetCode 772

python

1class Solution:
2    def calculate(self, s: str) -> int:
3        i = 0
4        
5        def parse():
6            nonlocal i
7            stack = []
8            op = '+'
9            num = 0
10            
11            while i < len(s):
12                c = s[i]
13                i += 1
14                
15                if c.isdigit():
16                    num = num * 10 + int(c)
17                elif c == '(':
18                    num = parse()
19                
20                # Check bounds or if char is an operator/parenthesis
21                # We check i >= len(s) inside because we just incremented i
22                if i >= len(s) or c in '+-*/)':
23                    if op == '+':
24                        stack.append(num)
25                    elif op == '-':
26                        stack.append(-num)
27                    elif op == '*':
28                        stack.append(stack.pop() * num)
29                    elif op == '/':
30                        # Python division floors (e.g., -3 // 2 = -2)
31                        # We need truncation towards zero (e.g., int(-3 / 2) = -1)
32                        stack.append(int(stack.pop() / num))
33                    
34                    op = c
35                    num = 0
36                    
37                    if c == ')':
38                        break
39            
40            return sum(stack)
41
42        return parse()

AI Coach Insight

For Basic Calculator III, focus on explaining why the structure works, not just naming Stack Patterns.

Complexity Analysis

Time Complexity: $O(N)$ We iterate through the string exactly once. Each character is processed a constant number of times (parsing, pushing to stack, popping). Even with recursion, the total operations are linear relative to the string length $N$ .

Space Complexity: $O(N)$ In the worst case (e.g., deeply nested parentheses like ((((...))))), the recursion stack depth can reach $O(N)$ . Additionally, the explicit stack used to store numbers can store up to $O(N)$ numbers in an expression like 1+2+3+4....

Related Patterns

Pattern: Stack Patterns
Subpattern: Stack - Expression Evaluation (RPN/Infix)

The problem fits the Stack - Expression Evaluation subpattern because mathematical expressions have a hierarchical structure defined by operator precedence and parentheses. A stack is the ideal data structure to hold temporary values (operands) while waiting for the correct operator precedence to be resolved.

Common Mistakes & FAQ

Why does my solution fail on negative division?
- Mistake: Using floor division (//) in Python.
- Concept Gap: Python's // operator rounds down to negative infinity, whereas C++/Java / operator truncates toward zero.
- Consequence: -3 // 2 returns -2 instead of -1. Use int(a / b) in Python.
Why do I get Wrong Answer on multi-digit numbers?
- Mistake: Treating num as a single digit (e.g., num = int(c)).
- Concept Gap: The input string contains characters. A number like "123" is read as '1', '2', '3'.
- Consequence: You must accumulate the number: num = num * 10 + digit.
Why does the loop terminate early or skip characters?
- Mistake: Incorrect index handling during recursion.
- Concept Gap: When returning from a recursive call for (...), the main loop's index must be advanced past the closing ).
- Consequence: If the index isn't shared (passed by reference or global), the outer loop will re-process the characters inside the parentheses.
Why does 1-2*3 evaluate to -3 instead of -5?
- Mistake: Evaluating 1-2 before 2*3.
- Concept Gap: Not using a stack to delay low-precedence operators.
- Consequence: The algorithm fails to respect order of operations. * and / must be executed immediately against the stack top, while + and - are pushed.

Practise Basic Calculator III and the problems around it

Basic Calculator III is a Stack Patterns problem, in the Expression Evaluation (RPN/Infix) subpattern. Once the approach above makes sense, write it yourself and then work through the neighbouring problems that reuse it.

Core insight

Basic Calculator III · Stack Patterns

Core Insight for Basic Calculator III

The key intuition for solving LeetCode 772 efficiently is to treat every expression (or sub-expression within parentheses) as a sequence of additions.

Reduction to Summation: We can treat subtraction A - B as A + (-B). This allows us to reduce the final step of any scope to simply summing up the stack.
Greedy Evaluation of High Precedence: When we encounter multiplication * or division /, we cannot wait. These operators bind tighter than + or -. We must pop the top value from our stack, perform the operation with the current number, and push the result back.
Delayed Evaluation of Low Precedence: When we encounter + or -, we push the previous number onto the stack and update our "pending operator." We do not evaluate the sum yet because a future * or / might need the current number.
Recursion for Parentheses: When we see a (, we are entering a sub-problem. This sub-problem is identical to the main problem but ends at ). We can recursively call our solver (utilizing the system call stack) to evaluate the sub-expression and return a single integer.

Visual Description: Imagine a stack representing the current "summation terms." If the expression is 2 + 3 * 4, the stack evolves as follows:

Read 2, operator is +. Push 2. Stack: [2].
Read 3, operator is *. Push 3. Stack: [2, 3].
Read 4. Previous operator was *. Pop 3, calculate 3 * 4 = 12. Push 12. Stack: [2, 12].
End of string. Sum the stack: 2 + 12 = 14.

The stack effectively isolates terms that are separated by + signs, while * and / merge terms immediately before they are finalized on the stack.

Deep dive

Full breakdown

Full reasoning, pitfalls, and implementation flow

LeetCode 772: Basic Calculator III Solution & Explanation

Problem Overview

TL;DR: The optimal solution uses recursion (implicit stack) to handle parentheses and an explicit stack to handle operator precedence, reducing the entire expression to a sum of integers.

This is a comprehensive interview question because it combines the logic of Basic Calculator I (parentheses) and Basic Calculator II (precedence).

Brute Force Approach for Basic Calculator III

Find the deepest set of parentheses.
Inside those parentheses, scan for * or /.
Perform the operation and create a new string with the result substituted.
Repeat until no * or / remain in that scope, then process + and -.
Remove the parentheses and repeat the process for the outer scope.

Pseudo-code:

text

while string has operators:
    find inner-most (...)
    solve multiplication/division inside
    solve addition/subtraction inside
    replace (...) with result in string
return result

Algorithm Strategy: Stack Patterns

We will implement a recursive solution where each level of recursion solves the expression within a pair of parentheses (or the entire string).

State Maintenance:

Stack: Stores integers to be summed at the end of the current scope.
Current Number (num): Builds the multi-digit integer currently being parsed.
Last Operator (op): Tracks the operator that precedes the num. Initialized to +.

Constraints & Boundaries:

The recursion handles the boundaries of ( and ).
The loop terminates when we reach the end of the string or a closing parenthesis ).

Logic:

Iterate through the string (or a shared index/iterator).
If the character is a digit, update num.
If the character is (, recursively call the function to evaluate the sub-expression. The result becomes our num.
If the character is an operator +, -, *, / or a closing parenthesis ) or the end of the string:
- Apply op to the current num and the top of the stack.
  - +: Push num.
  - -: Push -num.
  - *: Pop top, push top * num.
  - /: Pop top, push top / num (truncate toward zero).
- Update op to the current character.
- Reset num to 0.
If the character is ), break the loop.
Return the sum of elements in the stack.

Execution Flow

Let's trace 3 + 2 * ( 5 - 1 )

Start Main Scope: op = '+', stack = [].
Parse 3. op is +. Push 3. stack = [3]. Update op = '+'.
Parse 2. op is +. Push 2. stack = [3, 2]. Update op = '*'.
Encounter (. Recurse.
- Sub-Scope: op = '+', stack = [].
- Parse 5. op is +. Push 5. stack = [5]. Update op = '-'.
- Parse 1. op is -. Push -1. stack = [5, -1]. Update op = ')'.
- Encounter ). Break loop.
- Return sum of stack: 5 + (-1) = 4.
Back to Main Scope: The recursion returned 4. So effectively, num = 4.
Previous op was *. Pop 2 from stack. Calculate 2 * 4 = 8. Push 8. stack = [3, 8].
End of string.
Return sum of stack: 3 + 8 = 11.

Proof of Correctness

The algorithm's correctness relies on the invariant that the stack always contains terms that are ready to be added.

Base Case: Single numbers are pushed directly (or negated for subtraction).
Inductive Step (Precedence): Multiplicative operators (*, /) combine the current number with the last pushed number immediately. This ensures standard order of operations because the result is placed back on the stack as a single term, effectively treating a + b * c as a + (b*c).
Inductive Step (Parentheses): Recursion treats a sub-expression (...) as a single atomic number. Since the sub-expression is evaluated using the same logic, it correctly returns the evaluated integer, maintaining the structural integrity of the expression.

Thus, at the end of any scope, the stack contains a list of values [v1, v2, ..., vn] representing v1 + v2 + ... + vn.

Pattern Reuse Notes

The Stack - Expression Evaluation pattern is highly reusable. Understanding the mechanism of "delaying evaluation" using a stack is key for these related problems:

LeetCode 150: Evaluate Reverse Polish Notation: This is the "backend" of the calculator. It evaluates expressions already in postfix format.
LeetCode 224: Basic Calculator: Handles +, -, and (). It is a subset of 772. The logic is identical, though strictly handling signs is sufficient without a full multiplication stack.
LeetCode 227: Basic Calculator II: Handles +, -, *, / but no parentheses. This is exactly the logic inside one "scope" of Basic Calculator III.

Feynman coach · Before you peek at the code

Can you explain the core insight in your own words?

If you can explain it before reading code, you’re far more likely to recall it under interview pressure.

Loading your attempt…

Reference Implementation

C++ Solution for LeetCode 772

cpp

1class Solution {
2public:
3    int calculate(string s) {
4        int i = 0;
5        return parse(s, i);
6    }
7
8private:
9    int parse(const string& s, int& i) {
10        vector<long> stack;
11        char op = '+';
12        long num = 0;
13
14        while (i < s.length()) {
15            char c = s[i++];
16            
17            if (isdigit(c)) {
18                num = num * 10 + (c - '0');
19            } else if (c == '(') {
20                num = parse(s, i);
21            }
22
23            // If c is an operator, or ')', or we are at the end, process the last number
24            // Note: We check i < s.length() inside the loop, but need to process
25            // the last number if we hit the end.
26            // A trick is to process if !isdigit(c) and c != ' ' or i == s.length()
27            // However, since we incremented i, we look at the previous char logic implicitly.
28            // Let's stick to the standard check:
29            if ((!isdigit(c) && c != ' ') || i == s.length()) { // Check current char (which is s[i-1] now due to i++)
30                 // Wait, simpler logic: check the char we just read
31                 // If the char we just read was '(', we already recursed and have num.
32                 // If it was ')', we process and break.
33            }
34            
35            // Correct logic flow for loop:
36            // The check needs to happen if c is an operator or end of block
37            // Since we incremented i, c is the character we are processing.
38            if ((!isdigit(c) && c != ' ') || i > s.length()) { 
39                // Wait, logic correction: standard loop structure is better.
40                // Let's rewind to the standard clean pattern.
41            }
42        }
43        return 0; 
44    }
45};
46
47// Re-writing strictly clean C++ implementation
48class Solution {
49public:
50    int calculate(string s) {
51        int i = 0;
52        return helper(s, i);
53    }
54
55    int helper(const string& s, int& i) {
56        vector<long> stack;
57        char op = '+';
58        long num = 0;
59
60        while (i < s.length()) {
61            char c = s[i++];
62
63            if (isdigit(c)) {
64                num = num * 10 + (c - '0');
65            } else if (c == '(') {
66                num = helper(s, i);
67            }
68
69            // Process if operator, closing parenthesis, or end of string
70            if (i == s.length() || c == '+' || c == '-' || c == '*' || c == '/' || c == ')') {
71                if (op == '+') {
72                    stack.push_back(num);
73                } else if (op == '-') {
74                    stack.push_back(-num);
75                } else if (op == '*') {
76                    long last = stack.back();
77                    stack.pop_back();
78                    stack.push_back(last * num);
79                } else if (op == '/') {
80                    long last = stack.back();
81                    stack.pop_back();
82                    // Integer division truncates toward zero in C++
83                    stack.push_back(last / num);
84                }
85                
86                op = c;
87                num = 0;
88
89                if (c == ')') break;
90            }
91        }
92
93        long result = 0;
94        for (long n : stack) result += n;
95        return (int)result;
96    }
97};

Java Solution for LeetCode 772

java

1import java.util.*;
2
3class Solution {
4    private int i = 0; // Global index to track position across recursion
5
6    public int calculate(String s) {
7        i = 0;
8        return helper(s);
9    }
10
11    private int helper(String s) {
12        Stack<Integer> stack = new Stack<>();
13        char op = '+';
14        int num = 0;
15
16        while (i < s.length()) {
17            char c = s.charAt(i++);
18
19            if (Character.isDigit(c)) {
20                num = num * 10 + (c - '0');
21            } else if (c == '(') {
22                num = helper(s);
23            }
24
25            // If operator, ')', or end of string (handled by loop condition + internal check)
26            // We check if the character is meaningful for processing logic
27            if (i >= s.length() || c == '+' || c == '-' || c == '*' || c == '/' || c == ')') {
28                if (op == '+') {
29                    stack.push(num);
30                } else if (op == '-') {
31                    stack.push(-num);
32                } else if (op == '*') {
33                    stack.push(stack.pop() * num);
34                } else if (op == '/') {
35                    stack.push(stack.pop() / num);
36                }
37                
38                op = c;
39                num = 0;
40
41                if (c == ')') break;
42            }
43        }
44
45        int result = 0;
46        for (int n : stack) {
47            result += n;
48        }
49        return result;
50    }
51}

Python Solution for LeetCode 772

python

1class Solution:
2    def calculate(self, s: str) -> int:
3        i = 0
4        
5        def parse():
6            nonlocal i
7            stack = []
8            op = '+'
9            num = 0
10            
11            while i < len(s):
12                c = s[i]
13                i += 1
14                
15                if c.isdigit():
16                    num = num * 10 + int(c)
17                elif c == '(':
18                    num = parse()
19                
20                # Check bounds or if char is an operator/parenthesis
21                # We check i >= len(s) inside because we just incremented i
22                if i >= len(s) or c in '+-*/)':
23                    if op == '+':
24                        stack.append(num)
25                    elif op == '-':
26                        stack.append(-num)
27                    elif op == '*':
28                        stack.append(stack.pop() * num)
29                    elif op == '/':
30                        # Python division floors (e.g., -3 // 2 = -2)
31                        # We need truncation towards zero (e.g., int(-3 / 2) = -1)
32                        stack.append(int(stack.pop() / num))
33                    
34                    op = c
35                    num = 0
36                    
37                    if c == ')':
38                        break
39            
40            return sum(stack)
41
42        return parse()

Basic Calculator III

Core insight

Core Insight for Basic Calculator III

Full breakdown

LeetCode 772: Basic Calculator III Solution & Explanation

Problem Overview

Brute Force Approach for Basic Calculator III

Algorithm Strategy: Stack Patterns

Execution Flow

Proof of Correctness

Pattern Reuse Notes

Can you explain the core insight in your own words?

Reference Implementation

C++ Solution for LeetCode 772

Java Solution for LeetCode 772

Python Solution for LeetCode 772

Practise Basic Calculator III and the problems around it

Related problems

Basic Calculator III

Core insight

Core Insight for Basic Calculator III

Full breakdown

LeetCode 772: Basic Calculator III Solution & Explanation

Problem Overview

Brute Force Approach for Basic Calculator III

Algorithm Strategy: Stack Patterns

Execution Flow

Proof of Correctness

Pattern Reuse Notes

Can you explain the core insight in your own words?

Reference Implementation

C++ Solution for LeetCode 772

Java Solution for LeetCode 772

Python Solution for LeetCode 772

Practise Basic Calculator III and the problems around it

Related problems