Jump Game Problem in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Jump Game problem?

To determine if you can reach the last index of an array starting from the first index, given that each element represents the maximum jump length from that position.

beginner

In the Jump Game problem, what does each element in the input array represent?

Each element represents the maximum number of steps you can jump forward from that position.

intermediate

What is a greedy approach to solve the Jump Game problem?

Keep track of the farthest reachable index while iterating through the array. If at any point the current index is beyond the farthest reachable index, return false. If you reach or pass the last index, return true.

intermediate

Why does the greedy approach work for the Jump Game problem?

Because at each step, you only need to know the farthest position you can reach. If you can keep moving forward without getting stuck, you can reach the end.

intermediate

What is the time complexity of the greedy solution for the Jump Game problem?

O(n), where n is the length of the input array, because it requires a single pass through the array.

In the Jump Game problem, what does a zero in the array mean?

A You can jump exactly one step

B You can jump any number of steps from that position

C You must jump backward

D You cannot jump forward from that position

A zero means you cannot move forward from that position, which might block progress if you cannot jump over it.

What should you do if your current index is greater than the farthest reachable index in the greedy approach?

A Return false

B Return true

C Reset the farthest reachable index

D Continue iterating

If the current index is beyond the farthest reachable index, it means you cannot reach this position, so return false.

What is the initial value of the farthest reachable index in the greedy solution?

A 1

B Length of array

C 0

D -1

You start at index 0, so the farthest reachable index initially is 0.

If the input array is [2,3,1,1,4], can you reach the last index?

A Yes

B No

C Only if you jump backward

D Not enough information

You can jump from index 0 to 1 or 2, then continue to the last index.

What is the space complexity of the greedy solution for the Jump Game problem?

A O(n)

B O(1)

C O(log n)

D O(n^2)

The greedy solution uses only a few variables, so space complexity is constant O(1).

Explain how the greedy algorithm solves the Jump Game problem step-by-step.

Think about tracking the farthest position you can jump to as you move forward.

Describe a scenario where you cannot reach the last index in the Jump Game problem.

Consider what happens if you get stuck at a position with zero jump length.

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 according to the values.

B. The maximum jump length you can make from any position.

C. The minimum number of jumps needed 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 start at the first index and reach 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 the number of jumps or maximum jump length.

3. Final Answer:

   If you can reach the last index starting from the first index by jumping according to the values. -> 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 needed.
• Thinking about maximum jump length only.
• Assuming sum of jumps matters.

2. Which of the following C# code snippets correctly declares an array for the Jump Game input?

easy

A. int jumps[] = new int[5] {2, 3, 1, 1, 4};

B. int jumps = {2, 3, 1, 1, 4};

C. int[] jumps = new int[] {2, 3, 1, 1, 4};

D. int[] jumps = (2, 3, 1, 1, 4);

Solution

1. Step 1: Check array declaration syntax in C#

   In C#, arrays are declared with type[], then initialized with new type[] and values inside braces.

2. Step 2: Validate each option

   int[] jumps = new int[] {2, 3, 1, 1, 4}; uses correct syntax: int[] jumps = new int[] {2, 3, 1, 1, 4}; Options A, B, and D have syntax errors or invalid declarations.

3. Final Answer:

   int[] jumps = new int[] {2, 3, 1, 1, 4}; -> Option C

4. Quick Check:

   C# array syntax = int[] jumps = new int[] {2, 3, 1, 1, 4}; [OK]

Hint: Use 'int[] name = new int[] {values};' for arrays in C#. [OK]

Common Mistakes:

• Omitting 'new int[]' when initializing arrays.
• Using parentheses instead of braces for array values.
• Declaring arrays like in other languages (e.g., C/C++).

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

public static bool CanJump(int[] nums) {
    int maxReach = 0;
    for (int i = 0; i < nums.Length; i++) {
        if (i > maxReach) return false;
        maxReach = Math.Max(maxReach, i + nums[i]);
    }
    return true;
}

Console.WriteLine(CanJump(new int[] {3, 2, 1, 0, 4}));

medium

A. Runtime exception

B. True

C. Compile-time error

D. False

Solution

1. Step 1: Trace the loop and maxReach updates

   Start with maxReach=0. At 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 (4 > maxReach=3), return false.

2. Step 2: Understand the return value

   Since index 4 is not reachable, the function returns false.

3. Final Answer:

   False -> Option D

4. Quick Check:

   Index 4 unreachable = False [OK]

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

Common Mistakes:

• Assuming all arrays are reachable.
• Not updating maxReach correctly.
• Confusing true/false return values.

4. Identify the bug in this C# Jump Game code snippet:

public static bool CanJump(int[] nums) {
    int maxReach = 0;
    for (int i = 0; i <= nums.Length; i++) {
        if (i > maxReach) return false;
        maxReach = Math.Max(maxReach, i + nums[i]);
    }
    return true;
}

medium

A. The loop condition should be i < nums.Length, not i <= nums.Length.

B. maxReach should be initialized to -1 instead of 0.

C. The return type should be int, not bool.

D. The Math.Max function is used incorrectly.

Solution

1. Step 1: Analyze the for loop condition

   The loop runs from i=0 to i <= nums.Length, which means it tries to access nums[nums.Length], causing an index out of range error.

2. Step 2: Correct the loop boundary

   The loop should run while i < nums.Length to avoid accessing beyond the array bounds.

3. Final Answer:

   The loop condition should be i < nums.Length, not i <= nums.Length. -> Option A

4. Quick Check:

   Array index out of range fixed by loop condition [OK]

Hint: Use i < length for array loops to avoid out-of-bounds. [OK]

Common Mistakes:

• Using <= instead of < in loop conditions.
• Ignoring array index limits.
• Misunderstanding return types.

5. Given the array int[] nums = {2, 0, 2, 0, 1}, which modification to the Jump Game algorithm below will correctly determine if the last index is reachable?

public static bool CanJump(int[] nums) {
    int maxReach = 0;
    for (int i = 0; i < nums.Length; i++) {
        if (i > maxReach) return false;
        maxReach = Math.Max(maxReach, i + nums[i]);
    }
    return true;
}

hard

A. Add a check to skip indices where nums[i] == 0.

B. No modification needed; the current algorithm works correctly.

C. Change the loop to iterate backwards from the end to the start.

D. Return false immediately if maxReach does not increase in an iteration.

Solution

1. Step 1: Trace the algorithm with the given array

   Start maxReach=0. i=0: maxReach=max(0,0+2)=2. i=1 (1 <= 2): maxReach=max(2,1+0)=2. i=2 (2 <= 2): maxReach=max(2,2+2)=4. i=3 (3 <= 4): maxReach=max(4,3+0)=4. i=4 (4 <= 4): maxReach=max(4,4+1)=5. Loop ends, return true.

2. Step 2: Confirm algorithm correctness

   The algorithm correctly identifies the last index is reachable without modification.

3. Final Answer:

   No modification needed; the current algorithm works correctly. -> Option B

4. Quick Check:

   Algorithm handles zeros and reachability correctly [OK]

Hint: Test with zeros; algorithm handles them if maxReach updates properly. [OK]

Common Mistakes:

• Thinking zeros always block progress.
• Adding unnecessary checks for zero values.
• Changing loop direction without reason.