What is the time complexity of The Score of Students Solving Math Expression?

The optimal solution has a time complexity of O(n^3 * 1000).

What is the space complexity of The Score of Students Solving Math Expression?

The space complexity is O(n^2 * 1000).

What data structures are used to solve The Score of Students Solving Math Expression?

This problem uses Array, Hash Table, Math, String, Dynamic Programming, Stack, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving The Score of Students Solving Math Expression?

Insight 1: Dynamic Programming is essential to generate all possible results by evaluating the expression with different operator precedence, while respecting parentheses (even if implicit). The constraint on the maximum result (1000) helps to prune computations. Insight 2: Correctly evaluating the original expression based on the standard order of operations (multiplication before addition) is crucial.

Which companies ask LeetCode 2019. The Score of Students Solving Math Expression in interviews?

This problem is frequently asked at Flipkart.

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The Score of Students Solving Math Expression - LeetCode 2019 Solution

The Score of Students Solving Math Expression - Complete Solution Guide

The Score of Students Solving Math Expression is LeetCode problem 2019, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s that contains digits 0-9 , addition symbols '+' , and multiplication symbols '*' only , representing a valid math expression of single digit numbers (e.g., 3+5*2 ). This expression was given to n elementary school students. The students were instructed to get the answer of the expression by following this order of operations : Compute multiplication , reading from left to right ; Then, Compute addition , reading from left to right . You are given an integer array answers

Detailed Explanation

The problem asks us to evaluate the scores of students who solved a math expression. The expression contains single-digit numbers, addition '+', and multiplication '*' operations. Students were asked to calculate the result following the standard order of operations (multiplication before addition, from left to right). We're given the expression string `s` and an array `answers` containing students' submissions. The scoring is as follows: * 5 points if the answer matches the correct result of the expression. * 2 points if the answer can be obtained by performing operations in a different order, but with correct arithmetic. This means the student arrived at a correct numerical value through a different order of operations. * 0 points otherwise. The goal is to calculate the total score of all students, considering their answers and the correct result, along with all possible (incorrect order) results.

Solution Approach

The solution uses Dynamic Programming (DP) to generate all possible answers for the math expression with various operator precedence. Then, it calculates the correct answer following the proper order of operations. Finally, it iterates through the student's answers, assigning points based on whether the answer matches the correct result or is present in the set of possible, albeit incorrect-order, results.

Step-by-Step Algorithm

Step 1: Parse the input string `s` to separate numbers and operators.
Step 2: Implement a recursive DP function `dp(i, j)` that returns a set of all possible results for the substring `s[i:j+1]`. The base case is when i == j, in which case it returns a set containing only nums[i]. The DP uses memoization to avoid redundant calculations.
Step 3: Inside `dp(i, j)`, iterate from `k = i` to `j - 1` to split the expression at each operator. Recursively compute the possible results for the left and right subexpressions using `dp(i, k)` and `dp(k + 1, j)`. Combine the results from both subexpressions based on the operator `ops[k]` to create new possible results, adding them to the set. Only consider results <= 1000.
Step 4: Evaluate the expression correctly using standard order of operations (multiplication then addition). This typically uses an approach similar to expression parsing or a stack-based evaluation.
Step 5: Iterate through the `answers` array. For each answer, add 5 to the total score if it's equal to the correct answer. Otherwise, if the answer is present in the set of possible answers (calculated using DP), add 2 to the total score.
Step 6: Return the total score.

Key Insights

Insight 1: Dynamic Programming is essential to generate all possible results by evaluating the expression with different operator precedence, while respecting parentheses (even if implicit). The constraint on the maximum result (1000) helps to prune computations.
Insight 2: Correctly evaluating the original expression based on the standard order of operations (multiplication before addition) is crucial.
Insight 3: Using a set to store possible answers avoids duplicate computations and improves efficiency when searching for the existence of a student's answer within the set of possible answers.

Complexity Analysis

Time Complexity: O(n^3 * 1000)

Space Complexity: O(n^2 * 1000)

Topics

This problem involves: Array, Hash Table, Math, String, Dynamic Programming, Stack, Memoization.

Companies

Asked at: Flipkart.

The Score of Students Solving Math Expression | LeetCode 2019

Hard

Array Hash Table Math String Dynamic Programming Stack Memoization

Time: O(n^3 * 1000)Space: O(n^2 * 1000)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int scoreOfStudents(String s, int[] answers) {
        int n_str = s.length();
        List<Integer> nums = new ArrayList<>();
        List<Character> ops = new ArrayList<>();

        for (int i = 0; i < n_str; i += 2) {
            nums.add(Character.getNumericValue(s.charAt(i)));
        }

        for (int i = 1; i < n_str; i += 2) {
            ops.add(s.charAt(i));
        }

        int n = nums.size();
        Map<String, Set<Integer>> memo = new HashMap<>();

        
        //Override the dp function in Python
        class Helper{
            public Set<Integer> dp(int i, int j){
                String key = i + "," + j;
                if (memo.containsKey(key)) {
                    return memo.get(key);
                }

                if (i == j) {
                    Set<Integer> res = new HashSet<>();
                    res.add(nums.get(i));
                    memo.put(key, res);
                    return res;
                }

                Set<Integer> res_set = new HashSet<>();
                for (int k = i; k < j; k++) {
                    Set<Integer> left_res = dp(i, k);
                    Set<Integer> right_res = dp(k + 1, j);
                    char op = ops.get(k);

                    for (int v1 : left_res) {
                        for (int v2 : right_res) {
                            int res;
                            if (op == '+') {
                                res = v1 + v2;
                            } else {
                                res = v1 * v2;
                            }

                            if (res <= 1000) {
                                res_set.add(res);
                            }
                        }
                    }
                }

                memo.put(key, res_set);
                return res_set;
            }
        }
        
        Helper helper = new Helper();
        Set<Integer> possible_answers = helper.dp(0, n - 1);
        
        int correct_answer = evaluateExpression(s);

        int total_score = 0;
        Map<Integer, Integer> answer_counts = new HashMap<>();
        for (int ans : answers) {
            answer_counts.put(ans, answer_counts.getOrDefault(ans, 0) + 1);
        }

        for (Map.Entry<Integer, Integer> entry : answer_counts.entrySet()) {
            int ans = entry.getKey();
            int count = entry.getValue();

            if (ans == correct_answer) {
                total_score += 5 * count;
            } else if (possible_answers.contains(ans)) {
                total_score += 2 * count;
            }
        }

        return total_score;
    }

    private int evaluateExpression(String s) {
        return new Object() {
            int pos = -1, ch;

            void nextChar() {
                ch = (++pos < s.length()) ? s.charAt(pos) : -1;
            }

            boolean eat(int charToEat) {
                while (ch == ' ') nextChar();
                if (ch == charToEat) {
                    nextChar();
                    return true;
                }
                return false;
            }

            int parse() {
                nextChar();
                int x = parseExpression();
                if (pos < s.length()) throw new RuntimeException("Unexpected: " + (char)ch);
                return x;
            }

            // Grammar:
            // expression = term + expression | term
            // term = factor * term     | factor
            // factor = integer | ( expression )
            int parseExpression() {
                int x = parseTerm();
                for (;;) {
                    if (eat('+')) x += parseTerm(); // addition
                    else return x;
                }
            }

            int parseTerm() {
                int x = parseFactor();
                for (;;) {
                    if (eat('*')) x *= parseFactor(); // multiplication
                    else return x;
                }
            }

            int parseFactor() {
                if (eat('(')) { // parentheses
                    int x = parseExpression();
                    eat(')');
                    return x;
                } else if (ch >= '0' && ch <= '9') { // numbers
                    int x = 0;
                    while (ch >= '0' && ch <= '9') {
                        x = x * 10 + (ch - '0');
                        nextChar();
                    }
                    return x;
                } else {
                    throw new RuntimeException("Unexpected: " + (char)ch);
                }
            }
        }.parse();
    }
}

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Problem Info

Description

Problem 2019: The Score of Students Solving Math Expression - Detailed explanation and solution approaches.

Topics

Array Hash Table Math String Dynamic Programming Stack Memoization

Companies

Flipkart

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2019

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Parse the input string `s` to separate numbers and operators.
2
Step 2: Implement a recursive DP function `dp(i, j)` that returns a set of all possible results for the substring `s[i:j+1]`. The base case is when i == j, in which case it returns a set containing only nums[i]. The DP uses memoization to avoid redundant calculations.
3
Step 3: Inside `dp(i, j)`, iterate from `k = i` to `j - 1` to split the expression at each operator. Recursively compute the possible results for the left and right subexpressions using `dp(i, k)` and `dp(k + 1, j)`. Combine the results from both subexpressions based on the operator `ops[k]` to create new possible results, adding them to the set. Only consider results <= 1000.
4
Step 4: Evaluate the expression correctly using standard order of operations (multiplication then addition). This typically uses an approach similar to expression parsing or a stack-based evaluation.
5
Step 5: Iterate through the `answers` array. For each answer, add 5 to the total score if it's equal to the correct answer. Otherwise, if the answer is present in the set of possible answers (calculated using DP), add 2 to the total score.
6
Step 6: Return the total score.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^3 * 1000)

💾

Space Complexity

O(n^2 * 1000)

📝

Why?

* The DP has a nested loop structure determined by substring length (O(N^2)) and the internal loops for splitting the expression (O(N)). * For each sub-expression we are generating all the possible sums and products of the possible evaluation of the sub-expression, which in the worst case could be up to O(K^2) since each subexpression result is capped at 1000, giving an overall O(N^3*K^2) complexity. However, given K is a constraint within the problem, we consider this a constant giving an O(N^3) solution. Storing possible answers takes O(N^2 * K) space. Because all numbers are single digits and the maximum result is capped at 1000, intermediate results are also bounded by 1000.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Shunting Yard Algorithm and Expression Tree

Convert the infix expression to postfix (Reverse Polish Notation) using the Shunting Yard algorithm. Then, build an expression tree from the postfix notation. Evaluate the tree in standard order to get the correct answer. Use the expression tree to explore alternate evaluation orders (by manipulating the tree structure) to find the set of possible answers.

Time: O(N) for Shunting Yard and tree construction, but potentially exponential for exploring all possible evaluation orders. Worst case O(N!)Space: O(N) for the stack and the expression tree.

Trade-offs: While Shunting Yard is efficient for the standard evaluation, generating all possible evaluation orders from the expression tree could be very inefficient (exponential) and likely to exceed time limits. DP provides a better trade-off between time and space complexity for this problem.

✨

Key Insights

1Insight 1: Dynamic Programming is essential to generate all possible results by evaluating the expression with different operator precedence, while respecting parentheses (even if implicit). The constraint on the maximum result (1000) helps to prune computations.
2Insight 2: Correctly evaluating the original expression based on the standard order of operations (multiplication before addition) is crucial.
3Insight 3: Using a set to store possible answers avoids duplicate computations and improves efficiency when searching for the existence of a student's answer within the set of possible answers.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Expression with only one number (e.g., "5"). The DP should correctly handle this base case and return a set containing only that number.
Edge case 2: Expression with no multiplication. The correct evaluation should handle the addition properly. The DP should still work, exploring all possible orders of addition.
Edge case 3: Student gives the correct answer but could have arrived at it through incorrect operations. They still get 5 points.
Edge case 4: The expression is very long and contains a mix of additions and multiplications that causes the intermediate value in the DP to exceed the 1000 limit. The pruning condition 'res <= 1000' addresses this edge case by avoiding unnecessary computations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly evaluating the original expression. Forgetting the order of operations (multiplication before addition) will lead to the wrong 'correct answer' and an incorrect score.

✗Common mistake 2: Not memoizing the DP function. Without memoization, the DP function will recalculate the same subproblems multiple times, leading to exponential time complexity and exceeding time limits.

✗Common mistake 3: Failing to properly prune the intermediate results using the 'res <= 1000' condition. Ignoring this condition will lead to unnecessary calculations and potentially exceeding memory limits or the time limit.

✗Common mistake 4: Incorrectly using a list or array instead of a set. Using an array to store possible answers can be much slower as searching becomes O(n).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Compiler Design: Evaluating expressions in compilers and interpreters, where understanding operator precedence and generating different possible evaluation paths is relevant.

🏦Real-world scenario 2: Game Development: Scoring systems in games where a player might achieve a score through different combinations of actions. The system needs to determine if a player's score is valid even if the actions weren't performed in the optimal order.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "The Score of Students Solving Math Expression"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?