Why Jump Game Problem in DSA Go?

The Big Idea

What if you could skip all the guesswork and instantly know if you can reach the end?

The Scenario

Imagine you are playing a board game where each square tells you how many steps you can jump forward. You want to know if you can reach the last square starting from the first one. Doing this by checking every possible jump manually is like trying to guess the right moves without a plan.

The Problem

Manually trying every jump is slow and confusing. You might miss some paths or waste time checking impossible jumps. It's easy to get stuck or repeat the same checks over and over, making the game frustrating and error-prone.

The Solution

The Jump Game Problem uses a smart way to check if you can reach the end by keeping track of the farthest place you can jump so far. This way, you don't try every path but only focus on the best jumps, saving time and effort.

Before vs After

✗ Before

for i := 0; i < len(jumps); i++ {
    for j := 1; j <= jumps[i]; j++ {
        // check if can jump from i to i+j
    }
}

✓ After

maxReach := 0
for i := 0; i < len(jumps); i++ {
    if i > maxReach {
        return false
    }
    if maxReach < i+jumps[i] {
        maxReach = i + jumps[i]
    }
}
return true

What It Enables

This approach lets you quickly decide if the last square is reachable without wasting time on impossible jumps.

Real Life Example

Think of planning a road trip where each city sign tells you how far you can drive next. Instead of checking every route, you track the farthest city you can reach to decide if the destination is possible.

Key Takeaways

Manual checking of all jumps is slow and confusing.

Tracking the farthest reachable position simplifies the problem.

This method quickly tells if the end is reachable or not.

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 by jumping within allowed steps

B. The maximum jump length you can make 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 you can move from the first index to the last index by jumping forward within the allowed steps at each position.

2. Step 2: Identify the key question

   The key question is whether the last index is reachable, not how many jumps or maximum jump length.

3. Final Answer:

   If you can reach the last index starting from the first index by jumping within allowed steps -> Option A

4. Quick Check:

   Jump Game = Reach last index? [OK]

Hint: Focus on reachability, not jump count or max jump length [OK]

Common Mistakes:

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

2. Which of the following Go code snippets correctly declares an integer slice representing jump lengths for the Jump Game?

easy

A. jumps := [2, 3, 1, 1, 4]

B. jumps := (2, 3, 1, 1, 4)

C. jumps := int{2, 3, 1, 1, 4}

D. jumps := []int{2, 3, 1, 1, 4}

Solution

1. Step 1: Recall Go slice declaration syntax

   In Go, slices are declared with []type and curly braces for values, e.g., []int{...}.

2. Step 2: Check each option

   Only jumps := []int{2, 3, 1, 1, 4} uses correct slice syntax. The others use array syntax, invalid int{}, or parentheses.

3. Final Answer:

   jumps := []int{2, 3, 1, 1, 4} -> Option D

4. Quick Check:

   Slice declaration = []int{values} [OK]

Hint: Use []int{...} to declare integer slices in Go [OK]

Common Mistakes:

• Using square brackets without type for slices
• Using parentheses instead of braces
• Confusing arrays and slices syntax

3. What is the output of the following Go code snippet for the Jump Game problem?

package main

import "fmt"

func canJump(nums []int) bool {
    maxReach := 0
    for i := 0; i < len(nums); i++ {
        if i > maxReach {
            return false
        }
        if i+nums[i] > maxReach {
            maxReach = i + nums[i]
        }
    }
    return true
}

func main() {
    nums := []int{2, 3, 1, 1, 4}
    fmt.Println(canJump(nums))
}

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Trace maxReach updates

   Start maxReach=0. At i=0, maxReach=max(0,0+2)=2. i=1 ≤ 2, maxReach=max(2,1+3)=4. i=2 ≤ 4, maxReach=max(4,2+1)=4. i=3 ≤ 4, maxReach=max(4,3+1)=4. i=4 ≤ 4, maxReach=max(4,4+4)=8.

2. Step 2: Check if any index is unreachable

   All i ≤ maxReach, so no false return. Loop ends, returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   All indices reachable = true [OK]

Hint: Track max reachable index; if current index exceeds it, return false [OK]

Common Mistakes:

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

4. Identify the bug in this Go function for the Jump Game problem and select the fix:

func canJump(nums []int) bool {
    maxReach := 0
    for i := 0; i <= len(nums); i++ {
        if i > maxReach {
            return false
        }
        if i+nums[i] > maxReach {
            maxReach = i + nums[i]
        }
    }
    return true
}

medium

A. Remove the if condition checking i > maxReach

B. Initialize maxReach to 1 instead of 0

C. Change loop condition to i < len(nums)

D. Change return false to return true inside the loop

Solution

1. Step 1: Analyze loop boundary

   The loop runs with i <= len(nums), which causes index out of range error at i == len(nums).

2. Step 2: Fix loop condition

   Changing to i < len(nums) prevents out of range access and fixes the bug.

3. Final Answer:

   Change loop condition to i < len(nums) -> Option C

4. Quick Check:

   Loop index must be less than slice length [OK]

Hint: Use i < len(slice) to avoid index out of range errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring runtime panic from out of range
• Changing unrelated parts of code

5. Given the array nums := []int{3, 2, 1, 0, 4}, what is the result of applying the Jump Game algorithm to check if the last index is reachable?

hard

A. true

B. false

C. Runtime error due to zero jump

D. Depends on starting index

Solution

1. Step 1: Trace maxReach through the array

   Start maxReach=0. i=0: maxReach=max(0,0+3)=3. i=1 ≤ 3: maxReach=max(3,1+2)=3. i=2 ≤ 3: maxReach=max(3,2+1)=3. i=3 ≤ 3: maxReach=max(3,3+0)=3. i=4 > 3, so return false.

2. Step 2: Understand why last index is unreachable

   At index 3, jump length is 0, so maxReach does not increase. Index 4 is beyond maxReach, unreachable.

3. Final Answer:

   false -> Option B

4. Quick Check:

   Zero jump blocks progress = false [OK]

Hint: If maxReach stops before last index, return false [OK]

Common Mistakes:

• Assuming zero jump can be bypassed
• Ignoring unreachable last index
• Thinking runtime error occurs