Jump Game Problem in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if we can jump from the first position.

DSA Javascript

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i <= [1]; i++) {
    if (i > maxReach) return false;
    maxReach = Math.max(maxReach, i + nums[i]);
  }
  return true;
}

Drag options to blanks, or click blank then click option'

Anums.length - 1

Bnums.length

Cnums.length + 1

Dnums.length - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums.length instead of nums.length - 1 causes out-of-bound errors.

Looping too far or too short misses checking the last position.

We loop until the last index, which is nums.length - 1, to check if we can reach the end.

2fill in blank

medium

Complete the code to update the maximum reachable index.

DSA Javascript

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i <= nums.length - 1; i++) {
    if (i > maxReach) return false;
    maxReach = Math.max(maxReach, i [1] nums[i]);
  }
  return true;
}

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition changes the logic.

Forgetting to update maxReach properly causes wrong results.

We add i and nums[i] to find the farthest index reachable from position i.

3fill in blank

hard

Fix the error in the loop condition to avoid out-of-bounds errors.

DSA Javascript

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i [1] nums.length; i++) {
    if (i > maxReach) return false;
    maxReach = Math.max(maxReach, i + nums[i]);
  }
  return true;
}

Drag options to blanks, or click blank then click option'

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of range errors.

Using '>' or '>=' reverses the loop logic.

The loop should run while i is less than nums.length to avoid going out of bounds.

4fill in blank

hard

Fill both blanks to return false when the current index is unreachable and to update maxReach correctly.

DSA Javascript

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i < nums.length; i++) {
    if (i [1] maxReach) return false;
    maxReach = Math.max(maxReach, i [2] nums[i]);
  }
  return true;
}

Drag options to blanks, or click blank then click option'

A>

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' in the if condition reverses the logic.

Using subtraction instead of addition to update maxReach.

If current index i is greater than maxReach, we cannot reach it, so return false. Update maxReach by adding i and nums[i].

5fill in blank

hard

Fill all three blanks to create a function that returns true if the last index is reachable, using a loop and maxReach variable.

DSA Javascript

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i [1] nums.length; i++) {
    if (i [2] maxReach) return false;
    maxReach = Math.max(maxReach, i [3] nums[i]);
  }
  return true;
}

Drag options to blanks, or click blank then click option'

A<

B>

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' instead of '<' in loop condition causes errors.

Using '<' instead of '>' in if condition reverses logic.

Using '-' instead of '+' to update maxReach.

Loop while i is less than nums.length. If i is greater than maxReach, return false. Update maxReach by adding i and nums[i].

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