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2019. The Score of Students Solving Math Expression

Description

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:

  1. Compute multiplication, reading from left to right; Then,
  2. Compute addition, reading from left to right.

You are given an integer array answers of length n, which are the submitted answers of the students in no particular order. You are asked to grade the answers, by following these rules:

  • If an answer equals the correct answer of the expression, this student will be rewarded 5 points;
  • Otherwise, if the answer could be interpreted as if the student applied the operators in the wrong order but had correct arithmetic, this student will be rewarded 2 points;
  • Otherwise, this student will be rewarded 0 points.

Return the sum of the points of the students.

 

Example 1:

Input: s = "7+3*1*2", answers = [20,13,42]
Output: 7
Explanation: As illustrated above, the correct answer of the expression is 13, therefore one student is rewarded 5 points: [20,13,42]
A student might have applied the operators in this wrong order: ((7+3)*1)*2 = 20. Therefore one student is rewarded 2 points: [20,13,42]
The points for the students are: [2,5,0]. The sum of the points is 2+5+0=7.

Example 2:

Input: s = "3+5*2", answers = [13,0,10,13,13,16,16]
Output: 19
Explanation: The correct answer of the expression is 13, therefore three students are rewarded 5 points each: [13,0,10,13,13,16,16]
A student might have applied the operators in this wrong order: ((3+5)*2 = 16. Therefore two students are rewarded 2 points: [13,0,10,13,13,16,16]
The points for the students are: [5,0,0,5,5,2,2]. The sum of the points is 5+0+0+5+5+2+2=19.

Example 3:

Input: s = "6+0*1", answers = [12,9,6,4,8,6]
Output: 10
Explanation: The correct answer of the expression is 6.
If a student had incorrectly done (6+0)*1, the answer would also be 6.
By the rules of grading, the students will still be rewarded 5 points (as they got the correct answer), not 2 points.
The points for the students are: [0,0,5,0,0,5]. The sum of the points is 10.

 

Constraints:

  • 3 <= s.length <= 31
  • s represents a valid expression that contains only digits 0-9, '+', and '*' only.
  • All the integer operands in the expression are in the inclusive range [0, 9].
  • 1 <= The count of all operators ('+' and '*') in the math expression <= 15
  • Test data are generated such that the correct answer of the expression is in the range of [0, 1000].
  • n == answers.length
  • 1 <= n <= 104
  • 0 <= answers[i] <= 1000

Solutions

Solution 1: Dynamic Programming (Interval DP)

First, we design a function $cal(s)$ to calculate the result of a valid mathematical expression that only contains single-digit numbers. The correct answer is $x = cal(s)$.

Let the length of the string $s$ be $n$, then the number of digits in $s$ is $m = \frac{n+1}{2}$.

We define $f[i][j]$ as the possible values of the result calculated by selecting the digits from the $i$-th to the $j$-th in $s$ (index starts from $0$). Initially, $f[i][i]$ represents the selection of the $i$-th digit, and the result can only be this digit itself, i.e., $f[i][i] = {s[i \times 2]}$ (the $i$-th digit maps to the character at index $i \times 2$ in the string $s$).

Next, we enumerate $i$ from large to small, and then enumerate $j$ from small to large. We need to find out the possible values of the results of the operation of all digits in the interval $[i, j]$. We enumerate the boundary point $k$ in the interval $[i, j]$, then $f[i][j]$ can be obtained from $f[i][k]$ and $f[k+1][j]$ through the operator $s[k \times 2 + 1]$. Therefore, we can get the following state transition equation:

$$ f[i][j] = \begin{cases} {s[i \times 2]}, & i = j \ \bigcup\limits_{k=i}^{j-1} {f[i][k] \otimes f[k+1][j]}, & i < j \end{cases} $$

Where $\otimes$ represents the operator, i.e., $s[k \times 2 + 1]$.

The possible values of the results of all digit operations in the string $s$ are $f[0][m-1]$.

Finally, we count the answer. We use an array $cnt$ to count the number of times each answer appears in the answer array $answers$. If the answer is equal to $x$, then this student gets $5$ points, otherwise if the answer is in $f[0][m-1]$, then this student gets $2$ points. Traverse $cnt$ to count the answer.

The time complexity is $O(n^3 \times M^2)$, and the space complexity is $O(n^2 \times M^2)$. Here, $M$ is the maximum possible value of the answer, and $n$ is the number of digits in the length of the string $s$.

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class Solution:
    def scoreOfStudents(self, s: str, answers: List[int]) -> int:
        def cal(s: str) -> int:
            res, pre = 0, int(s[0])
            for i in range(1, n, 2):
                if s[i] == "*":
                    pre *= int(s[i + 1])
                else:
                    res += pre
                    pre = int(s[i + 1])
            res += pre
            return res

        n = len(s)
        x = cal(s)
        m = (n + 1) >> 1
        f = [[set() for _ in range(m)] for _ in range(m)]
        for i in range(m):
            f[i][i] = {int(s[i << 1])}
        for i in range(m - 1, -1, -1):
            for j in range(i, m):
                for k in range(i, j):
                    for l in f[i][k]:
                        for r in f[k + 1][j]:
                            if s[k << 1 | 1] == "+" and l + r <= 1000:
                                f[i][j].add(l + r)
                            elif s[k << 1 | 1] == "*" and l * r <= 1000:
                                f[i][j].add(l * r)
        cnt = Counter(answers)
        ans = cnt[x] * 5
        for k, v in cnt.items():
            if k != x and k in f[0][m - 1]:
                ans += v << 1
        return ans

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class Solution {
    public int scoreOfStudents(String s, int[] answers) {
        int n = s.length();
        int x = cal(s);
        int m = (n + 1) >> 1;
        Set<Integer>[][] f = new Set[m][m];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < m; ++j) {
                f[i][j] = new HashSet<>();
            }
            f[i][i].add(s.charAt(i << 1) - '0');
        }
        for (int i = m - 1; i >= 0; --i) {
            for (int j = i; j < m; ++j) {
                for (int k = i; k < j; ++k) {
                    for (int l : f[i][k]) {
                        for (int r : f[k + 1][j]) {
                            char op = s.charAt(k << 1 | 1);
                            if (op == '+' && l + r <= 1000) {
                                f[i][j].add(l + r);
                            } else if (op == '*' && l * r <= 1000) {
                                f[i][j].add(l * r);
                            }
                        }
                    }
                }
            }
        }
        int[] cnt = new int[1001];
        for (int ans : answers) {
            ++cnt[ans];
        }
        int ans = 5 * cnt[x];
        for (int i = 0; i <= 1000; ++i) {
            if (i != x && f[0][m - 1].contains(i)) {
                ans += 2 * cnt[i];
            }
        }
        return ans;
    }

    private int cal(String s) {
        int res = 0, pre = s.charAt(0) - '0';
        for (int i = 1; i < s.length(); i += 2) {
            char op = s.charAt(i);
            int cur = s.charAt(i + 1) - '0';
            if (op == '*') {
                pre *= cur;
            } else {
                res += pre;
                pre = cur;
            }
        }
        res += pre;
        return res;
    }
}

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class Solution {
public:
    int scoreOfStudents(string s, vector<int>& answers) {
        int n = s.size();
        int x = cal(s);
        int m = (n + 1) >> 1;
        unordered_set<int> f[m][m];
        for (int i = 0; i < m; ++i) {
            f[i][i] = {s[i * 2] - '0'};
        }
        for (int i = m - 1; ~i; --i) {
            for (int j = i; j < m; ++j) {
                for (int k = i; k < j; ++k) {
                    for (int l : f[i][k]) {
                        for (int r : f[k + 1][j]) {
                            char op = s[k << 1 | 1];
                            if (op == '+' && l + r <= 1000) {
                                f[i][j].insert(l + r);
                            } else if (op == '*' && l * r <= 1000) {
                                f[i][j].insert(l * r);
                            }
                        }
                    }
                }
            }
        }
        int cnt[1001]{};
        for (int t : answers) {
            ++cnt[t];
        }
        int ans = 5 * cnt[x];
        for (int i = 0; i <= 1000; ++i) {
            if (i != x && f[0][m - 1].count(i)) {
                ans += cnt[i] << 1;
            }
        }
        return ans;
    }

    int cal(string& s) {
        int res = 0;
        int pre = s[0] - '0';
        for (int i = 1; i < s.size(); i += 2) {
            int cur = s[i + 1] - '0';
            if (s[i] == '*') {
                pre *= cur;
            } else {
                res += pre;
                pre = cur;
            }
        }
        res += pre;
        return res;
    }
};

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func scoreOfStudents(s string, answers []int) int {
    n := len(s)
    x := cal(s)
    m := (n + 1) >> 1
    f := make([][]map[int]bool, m)
    for i := range f {
        f[i] = make([]map[int]bool, m)
        for j := range f[i] {
            f[i][j] = make(map[int]bool)
        }
        f[i][i][int(s[i<<1]-'0')] = true
    }
    for i := m - 1; i >= 0; i-- {
        for j := i; j < m; j++ {
            for k := i; k < j; k++ {
                for l := range f[i][k] {
                    for r := range f[k+1][j] {
                        op := s[k<<1|1]
                        if op == '+' && l+r <= 1000 {
                            f[i][j][l+r] = true
                        } else if op == '*' && l*r <= 1000 {
                            f[i][j][l*r] = true
                        }
                    }
                }
            }
        }
    }
    cnt := [1001]int{}
    for _, v := range answers {
        cnt[v]++
    }
    ans := cnt[x] * 5
    for k, v := range cnt {
        if k != x && f[0][m-1][k] {
            ans += v << 1
        }
    }
    return ans
}

func cal(s string) int {
    res, pre := 0, int(s[0]-'0')
    for i := 1; i < len(s); i += 2 {
        cur := int(s[i+1] - '0')
        if s[i] == '+' {
            res += pre
            pre = cur
        } else {
            pre *= cur
        }
    }
    res += pre
    return res
}

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function scoreOfStudents(s: string, answers: number[]): number {
    const n = s.length;
    const cal = (s: string): number => {
        let res = 0;
        let pre = s.charCodeAt(0) - '0'.charCodeAt(0);
        for (let i = 1; i < s.length; i += 2) {
            const cur = s.charCodeAt(i + 1) - '0'.charCodeAt(0);
            if (s[i] === '+') {
                res += pre;
                pre = cur;
            } else {
                pre *= cur;
            }
        }
        res += pre;
        return res;
    };
    const x = cal(s);
    const m = (n + 1) >> 1;
    const f: Set<number>[][] = Array(m)
        .fill(0)
        .map(() =>
            Array(m)
                .fill(0)
                .map(() => new Set()),
        );
    for (let i = 0; i < m; ++i) {
        f[i][i].add(s[i << 1].charCodeAt(0) - '0'.charCodeAt(0));
    }
    for (let i = m - 1; i >= 0; --i) {
        for (let j = i; j < m; ++j) {
            for (let k = i; k < j; ++k) {
                for (const l of f[i][k]) {
                    for (const r of f[k + 1][j]) {
                        const op = s[(k << 1) + 1];
                        if (op === '+' && l + r <= 1000) {
                            f[i][j].add(l + r);
                        } else if (op === '*' && l * r <= 1000) {
                            f[i][j].add(l * r);
                        }
                    }
                }
            }
        }
    }
    const cnt: number[] = Array(1001).fill(0);
    for (const v of answers) {
        ++cnt[v];
    }
    let ans = cnt[x] * 5;
    for (let i = 0; i <= 1000; ++i) {
        if (i !== x && f[0][m - 1].has(i)) {
            ans += cnt[i] << 1;
        }
    }
    return ans;
}

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