3522. Calculate Score After Performing Instructions 

中文文档

Description

You are given two arrays, instructions and values, both of size n.

You need to simulate a process based on the following rules:

You start at the first instruction at index i = 0 with an initial score of 0.
If instructions[i] is "add":
- Add values[i] to your score.
- Move to the next instruction (i + 1).
If instructions[i] is "jump":
- Move to the instruction at index (i + values[i]) without modifying your score.

The process ends when you either:

Go out of bounds (i.e., i < 0 or i >= n), or
Attempt to revisit an instruction that has been previously executed. The revisited instruction is not executed.

Return your score at the end of the process.

Example 1:

Input: instructions = ["jump","add","add","jump","add","jump"], values = [2,1,3,1,-2,-3]

Output: 1

Explanation:

Simulate the process starting at instruction 0:

At index 0: Instruction is "jump", move to index 0 + 2 = 2.
At index 2: Instruction is "add", add values[2] = 3 to your score and move to index 3. Your score becomes 3.
At index 3: Instruction is "jump", move to index 3 + 1 = 4.
At index 4: Instruction is "add", add values[4] = -2 to your score and move to index 5. Your score becomes 1.
At index 5: Instruction is "jump", move to index 5 + (-3) = 2.
At index 2: Already visited. The process ends.

Example 2:

Input: instructions = ["jump","add","add"], values = [3,1,1]

Output: 0

Explanation:

Simulate the process starting at instruction 0:

At index 0: Instruction is "jump", move to index 0 + 3 = 3.
At index 3: Out of bounds. The process ends.

Example 3:

Input: instructions = ["jump"], values = [0]

Output: 0

Explanation:

Simulate the process starting at instruction 0:

At index 0: Instruction is "jump", move to index 0 + 0 = 0.
At index 0: Already visited. The process ends.

Constraints:

n == instructions.length == values.length
1 <= n <= 10⁵
instructions[i] is either "add" or "jump".
-10⁵ <= values[i] <= 10⁵

Solutions

Solution 1: Simulation

We can simulate the process based on the problem description.

Define a boolean array $\textit{vis}$ of length $n$ to record whether each instruction has been executed. Initially, all elements are set to $\text{false}$.

Then, starting from index $i = 0$, perform the following steps in a loop:

Set $\textit{vis}[i]$ to $\text{true}$.
If the first character of $\textit{instructions}[i]$ is 'a', add $\textit{value}[i]$ to the answer and increment $i$ by $1$. Otherwise, increment $i$ by $\textit{value}[i]$.

The loop continues until $i \lt 0$, $i \ge n$, or $\textit{vis}[i]$ is $\text{true}$.

Finally, return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{value}$.

Python3

class Solution:
    def calculateScore(self, instructions: List[str], values: List[int]) -> int:
        n = len(values)
        vis = [False] * n
        ans = i = 0
        while 0 <= i < n and not vis[i]:
            vis[i] = True
            if instructions[i][0] == "a":
                ans += values[i]
                i += 1
            else:
                i = i + values[i]
        return ans

Java

class Solution {
    public long calculateScore(String[] instructions, int[] values) {
        int n = values.length;
        boolean[] vis = new boolean[n];
        long ans = 0;
        int i = 0;

        while (i >= 0 && i < n && !vis[i]) {
            vis[i] = true;
            if (instructions[i].charAt(0) == 'a') {
                ans += values[i];
                i += 1;
            } else {
                i = i + values[i];
            }
        }

        return ans;
    }
}

C++

class Solution {
public:
    long long calculateScore(vector<string>& instructions, vector<int>& values) {
        int n = values.size();
        vector<bool> vis(n, false);
        long long ans = 0;
        int i = 0;

        while (i >= 0 && i < n && !vis[i]) {
            vis[i] = true;
            if (instructions[i][0] == 'a') {
                ans += values[i];
                i += 1;
            } else {
                i += values[i];
            }
        }

        return ans;
    }
};

Go

func calculateScore(instructions []string, values []int) (ans int64) {
	n := len(values)
	vis := make([]bool, n)
	i := 0
	for i >= 0 && i < n && !vis[i] {
		vis[i] = true
		if instructions[i][0] == 'a' {
			ans += int64(values[i])
			i += 1
		} else {
			i += values[i]
		}
	}
	return
}

TypeScript

function calculateScore(instructions: string[], values: number[]): number {
    const n = values.length;
    const vis: boolean[] = Array(n).fill(false);
    let ans = 0;
    let i = 0;

    while (i >= 0 && i < n && !vis[i]) {
        vis[i] = true;
        if (instructions[i][0] === 'a') {
            ans += values[i];
            i += 1;
        } else {
            i += values[i];
        }
    }

    return ans;
}