LeetCode #3522 — MEDIUM

Calculate Score After Performing Instructions

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

["jump","add","add","jump","add","jump"]
[2,1,3,1,-2,-3]

Example 2

["jump","add","add"]
[3,1,1]

Example 3

["jump"]
[0]

Step 02

Core Insight

What unlocks the optimal approach

Simulate the process step by step, following the rules for each instruction.
Use a data structure to track which instructions have already been executed to detect revisits.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3522: Calculate Score After Performing Instructions
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;
    }
}

// Accepted solution for LeetCode #3522: Calculate Score After Performing Instructions
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
}

# Accepted solution for LeetCode #3522: Calculate Score After Performing Instructions
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

// Accepted solution for LeetCode #3522: Calculate Score After Performing Instructions
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3522: Calculate Score After Performing Instructions
// 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;
//     }
// }

// Accepted solution for LeetCode #3522: Calculate Score After Performing Instructions
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.