Why Jump Game Problem in DSA Javascript?

The Big Idea

Can you jump through the path without getting stuck? Discover the smart way!

The Scenario

Imagine you are playing a board game where each square tells you how many steps you can jump forward. You want to reach the last square, but you have to decide carefully which squares to jump on.

The Problem

Trying every possible jump by hand is slow and confusing. You might jump too far or too short and get stuck. It's easy to lose track and make mistakes, especially if the board is long.

The Solution

The Jump Game Problem helps you quickly find out if you can reach the end by checking the best jumps step-by-step. It uses smart rules to avoid trying every jump, saving time and effort.

Before vs After

✗ Before

function canReachEnd(board) {
  for (let i = 0; i < board.length; i++) {
    // try all jumps from i
  }
  return false;
}

✓ After

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

What It Enables

This problem lets you quickly decide if you can jump through a path without checking every step manually.

Real Life Example

Think of planning a road trip where each city tells you how far you can drive next. You want to know if you can reach your destination without running out of fuel.

Key Takeaways

Manual checking of all jumps is slow and error-prone.

The Jump Game Problem uses a smart way to track the farthest reachable point.

This helps quickly decide if the end is reachable.

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