Jump Game Problem in DSA Javascript - Build from Scratch

Jump Game Problem

📖 Scenario: Imagine you are playing a video game where you need to jump across platforms to reach the end. Each platform lets you jump a certain maximum number of steps forward. You want to check if it's possible to reach the last platform starting from the first one.

🎯 Goal: Build a program that checks if you can jump from the first platform to the last platform using the maximum jump lengths given for each platform.

📋 What You'll Learn

Create an array called jumps with exact values [2, 3, 1, 1, 4]

Create a variable called maxReach and set it to 0

Use a for loop with variable i to iterate over the jumps array

Inside the loop, update maxReach to the maximum of maxReach and i + jumps[i]

If maxReach is less than i at any point before the last index, break the loop

After the loop, print true if maxReach is greater than or equal to the last index, otherwise print false

💡 Why This Matters

🌍 Real World

This problem models real-world scenarios like planning moves in games, robot path planning, or network packet forwarding where you need to check reachability.

💼 Career

Understanding this problem helps in software engineering roles that require problem-solving skills, especially in game development, robotics, and algorithm design.

1

Create the jumps array

Create an array called jumps with these exact values: [2, 3, 1, 1, 4]

DSA Javascript

# Your code here

Hint

Use const jumps = [2, 3, 1, 1, 4]; to create the array.

2

Create maxReach variable

Create a variable called maxReach and set it to 0

DSA Javascript

const jumps = [2, 3, 1, 1, 4];
# Your code here

Hint

Use let maxReach = 0; to create the variable.

3

Implement the jump check loop

Use a for loop with variable i to iterate over the jumps array. Inside the loop, update maxReach to the maximum of maxReach and i + jumps[i]. If maxReach is less than i at any point before the last index, break the loop.

DSA Javascript

const jumps = [2, 3, 1, 1, 4];
let maxReach = 0;
# Your code here

Hint

Use for (let i = 0; i < jumps.length; i++) and update maxReach inside the loop.

4

Print if last platform is reachable

After the loop, print true if maxReach is greater than or equal to the last index of jumps, otherwise print false

DSA Javascript

const jumps = [2, 3, 1, 1, 4];
let maxReach = 0;
for (let i = 0; i < jumps.length; i++) {
    if (maxReach < i) {
        break;
    }
    maxReach = Math.max(maxReach, i + jumps[i]);
}
# Your code here

Hint

Use console.log(maxReach >= jumps.length - 1); to print the result.

Practice

1. What does the Jump Game problem ask you to determine given an array of non-negative integers?

easy

A. If you can reach the last index starting from the first index

B. The maximum jump length from any position

C. The minimum number of jumps to reach the last index

D. The sum of all jump lengths in the array

Solution

1. Step 1: Understand the problem statement

   The Jump Game problem asks if it is possible to move from the first index to the last index by jumping according to the values in the array.

2. Step 2: Identify the goal

   The goal is to check if the last index is reachable, not the number of jumps or maximum jump length.

3. Final Answer:

   If you can reach the last index starting from the first index -> Option A

4. Quick Check:

   Jump Game = Reach last index [OK]

Hint: Focus on reachability, not jump counts or sums [OK]

Common Mistakes:

• Confusing reachability with minimum jumps
• Thinking about maximum jump length only
• Assuming sum of jumps matters

2. Which of the following is the correct syntax to check if you can jump from index i to index j in JavaScript, given an array nums?

easy

A. if (nums[i] >= j - i)

B. if (nums[j] <= i + nums[i])

C. if (j < nums[i])

D. if (j <= i + nums[i])

Solution

1. Step 1: Understand jump condition

   You can jump from index i to j if j is within the jump length from i, i.e., j ≤ i + nums[i].

2. Step 2: Match condition with options

   if (j <= i + nums[i]) correctly checks if j ≤ i + nums[i]. Other options compare wrong values or use incorrect operators.

3. Final Answer:

   if (j <= i + nums[i]) -> Option D

4. Quick Check:

   Jump condition = j ≤ i + nums[i] [OK]

Hint: Check if target index ≤ current index + jump length [OK]

Common Mistakes:

• Comparing nums[j] instead of j
• Using < instead of ≤
• Mixing indices and values incorrectly

3. What will be the output of the following JavaScript code?

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i < nums.length; i++) {
    if (i > maxReach) return false;
    maxReach = Math.max(maxReach, i + nums[i]);
  }
  return true;
}

console.log(canJump([2,3,1,1,4]));
console.log(canJump([3,2,1,0,4]));

medium

A. true false

B. true true

C. false true

D. false false

Solution

1. Step 1: Trace first input [2,3,1,1,4]

   maxReach updates as: i=0 maxReach=2, i=1 maxReach=4, i=2 maxReach=4, i=3 maxReach=4, i=4 maxReach=8. No i > maxReach, so returns true.

2. Step 2: Trace second input [3,2,1,0,4]

   maxReach updates: i=0 maxReach=3, i=1 maxReach=3, i=2 maxReach=3, i=3 maxReach=3, i=4 i=4 > maxReach=3 triggers return false.

3. Final Answer:

   true false -> Option A

4. Quick Check:

   First array reachable, second not [OK]

Hint: Track max reachable index; fail if current index exceeds it [OK]

Common Mistakes:

• Not updating maxReach correctly
• Returning true too early
• Ignoring the condition i > maxReach

4. Identify the bug in the following code that attempts to solve the Jump Game problem:

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i <= nums.length; i++) {
    if (i > maxReach) return false;
    maxReach = Math.max(maxReach, i + nums[i]);
  }
  return true;
}

medium

A. maxReach should be initialized to 1, not 0

B. The loop condition should be i < nums.length, not i <= nums.length

C. The return true statement should be inside the loop

D. The comparison should be i >= maxReach instead of i > maxReach

Solution

1. Step 1: Analyze loop boundary

   The loop runs from i = 0 to i <= nums.length, which means it accesses nums[nums.length], causing undefined access.

2. Step 2: Correct loop condition

   The loop should run while i < nums.length to avoid out-of-bounds errors.

3. Final Answer:

   The loop condition should be i < nums.length, not i <= nums.length -> Option B

4. Quick Check:

   Loop boundary off by one error [OK]

Hint: Check array indices carefully to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= instead of < in loops
• Misinitializing maxReach
• Placing return true inside loop

5. Given the array nums = [1,2,0,1,0,2,0], which of the following statements is true about reaching the last index using the Jump Game logic?

hard

A. You can reach the last index only if you jump exactly 2 steps each time

B. You can reach the last index because jumps can skip zeros if reachable

C. You cannot reach the last index because zeros block all paths

D. You can reach the last index only if you jump backward at some point

Solution

1. Step 1: Trace maxReach for [1,2,0,1,0,2,0]

   i=0, maxReach=1; i=1, maxReach=3; i=2, maxReach=3; i=3, maxReach=4; i=4, maxReach=4; i=5 >4, cannot reach index 6.

2. Step 2: Analyze blocking

   Zeros at indices 2,4 limit jumps from reachable positions (max=4), blocking paths to end.

3. Final Answer:

   You cannot reach the last index because zeros block all paths -> Option C

4. Quick Check:

   Max reach=4 < length-1=6 [OK]

Hint: Trace maxReach step-by-step; zeros block if they stall reach before end [OK]

Common Mistakes:

• Thinking zeros can always be skipped sufficiently
• Assuming jumps must be fixed length
• Believing backward jumps are allowed