Why Jump Game Problem in DSA C?

The Big Idea

What if you could know instantly if you can finish a tricky jumping game without trying every step?

The Scenario

Imagine you are playing a board game where you can jump forward a certain number of steps based on the number on the square you land on. You want to know if you can reach the last square starting from the first one.

The Problem

Trying every possible jump manually is slow and confusing. You might miss some paths or waste time checking impossible jumps. It's easy to get stuck or lose track of where you can jump next.

The Solution

The Jump Game Problem uses a smart way to check if you can reach the end by tracking the farthest place you can jump to as you move forward. This avoids checking every path and quickly tells you if the last square is reachable.

Before vs After

✗ Before

int canReachEnd(int* jumps, int size) {
    // Try all jumps manually
    for (int i = 0; i < size; i++) {
        // Check each jump possibility
    }
    return 0;
}

✓ After

int canReachEnd(int* jumps, int size) {
    int maxReach = 0;
    for (int i = 0; i < size; i++) {
        if (i > maxReach) return 0;
        if (i + jumps[i] > maxReach) maxReach = i + jumps[i];
    }
    return 1;
}

What It Enables

This concept lets you quickly decide if you can reach the end of the game without trying every jump, saving time and effort.

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 or roads.

Key Takeaways

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

Tracking the farthest reachable point simplifies the problem.

This method quickly tells if the last position is reachable.

Practice

1. What is the main goal of the Jump Game problem?

easy

A. Calculate the minimum jump length required to reach the last index

B. Determine if you can reach the last index from the first index by jumping

C. Count the total number of jumps needed to reach the end

D. Find the maximum jump length from the first index

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 forward using the values at each index as maximum jump lengths.

2. Step 2: Identify the goal

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

3. Final Answer:

   Determine if you can reach the last index from the first index by jumping -> Option B

4. Quick Check:

   Goal = Reach last index [OK]

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

Common Mistakes:

• Confusing maximum jump length with reachability
• Trying to count jumps instead of checking possibility
• Assuming jumps can go backward

2. Which of the following is the correct syntax to declare an integer array nums of size 5 in C for the Jump Game problem?

easy

A. int[] nums = new int[5];

B. int nums = [5];

C. int nums(5);

D. int nums[5];

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared by specifying the type, name, and size in square brackets, like int nums[5];.

2. Step 2: Check each option

   int nums[5]; matches correct C syntax. int nums = [5]; uses invalid assignment. int nums(5); uses parentheses which is incorrect. int[] nums = new int[5]; is Java syntax.

3. Final Answer:

   int nums[5]; -> Option D

4. Quick Check:

   C array syntax = int nums[5]; [OK]

Hint: Use type name[size] for arrays in C [OK]

Common Mistakes:

• Using Java or C++ style array declarations
• Using parentheses instead of square brackets
• Assigning array size with equals sign

3. What will be the output of the following C code snippet for the Jump Game problem?

int nums[] = {2, 3, 1, 1, 4};
int n = 5;
int maxReach = 0;
for (int i = 0; i < n; i++) {
    if (i > maxReach) {
        printf("false\n");
        return 0;
    }
    if (i + nums[i] > maxReach) {
        maxReach = i + nums[i];
    }
}
printf("true\n");

medium

A. true

B. false

C. Compilation error

D. Runtime error

Solution

1. Step 1: Trace the loop and maxReach updates

   Start with 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 loop completes without returning false

   Since i never exceeds maxReach, loop completes and prints "true".

3. Final Answer:

   true -> Option A

4. Quick Check:

   maxReach covers all indices = true [OK]

Hint: Check if current index exceeds max reachable index [OK]

Common Mistakes:

• Stopping early without updating maxReach
• Confusing index with jump length
• Assuming false if maxReach equals index

4. Identify the bug in the following C code snippet for the Jump Game problem:

int canJump(int* nums, int n) {
    int maxReach = 0;
    for (int i = 0; i <= n; i++) {
        if (i > maxReach) return 0;
        if (i + nums[i] > maxReach) maxReach = i + nums[i];
    }
    return 1;
}

medium

A. Return values should be swapped (1 for false, 0 for true)

B. maxReach should be initialized to -1

C. Loop condition should be i < n, not i <= n

D. nums array should be declared inside the function

Solution

1. Step 1: Analyze the loop condition

   The loop runs from i=0 to i<=n, which means it accesses nums[n], an out-of-bounds index causing undefined behavior.

2. Step 2: Correct the loop condition

   The loop should run while i < n to avoid accessing beyond the array.

3. Final Answer:

   Loop condition should be i < n, not i <= n -> Option C

4. Quick Check:

   Array index in bounds = i < n [OK]

Hint: Loop up to n-1 to avoid out-of-bounds access [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring array boundary checks
• Incorrect return values for boolean

5. Given the array nums = {3, 2, 1, 0, 4}, why does the standard greedy Jump Game algorithm return false for reachability?

hard

A. Because index 3 has a jump length of 0, blocking further progress

B. Because the array length is less than 5

C. Because the first element is too large

D. Because the last element is unreachable due to negative jumps

Solution

1. Step 1: Trace jumps from start

   Start at index 0 with jump length 3, maxReach = 3. Indices 1, 2, 3 are reachable. At index 3, jump length is 0, so maxReach stays 3.

2. Step 2: Check if last index is reachable

   Last index is 4, which is beyond maxReach 3, so it is unreachable, causing the algorithm to return false.

3. Final Answer:

   Because index 3 has a jump length of 0, blocking further progress -> Option A

4. Quick Check:

   Zero jump blocks path = false [OK]

Hint: Zero jump at reachable index blocks reaching the end [OK]

Common Mistakes:

• Ignoring zero jump blocking
• Assuming array length affects reachability
• Thinking negative jumps exist in problem