Jump Game Problem in DSA C# - Time & Space Complexity
We want to understand how the time needed to check if we can jump to the end grows as the input array gets bigger.
How does the number of steps change when the array length increases?
Analyze the time complexity of the following code snippet.
public 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;
}
This code checks if you can jump from the start to the end of the array, where each number tells the max jump length from that position.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: A single loop that goes through each element once.
- How many times: Exactly once for each element in the array.
As the array gets longer, the number of steps grows directly with its length.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 10 steps |
| 100 | About 100 steps |
| 1000 | About 1000 steps |
Pattern observation: The steps increase in a straight line as the input size grows.
Time Complexity: O(n)
This means the time to decide if you can reach the end grows in direct proportion to the array size.
[X] Wrong: "Because we check jumps from each position, the time is much bigger than the array length, maybe quadratic."
[OK] Correct: The code only moves forward once through the array without nested loops, so it does not repeat work for each element multiple times.
Understanding this helps you explain how to efficiently check reachability in arrays, a common skill in coding interviews.
"What if we changed the code to try all possible jumps from each position recursively? How would the time complexity change?"
Practice
Solution
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.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.Final Answer:
If you can reach the last index starting from the first index by jumping according to the values. -> Option AQuick Check:
Jump Game = Reach last index? [OK]
- Confusing reachability with minimum jumps needed.
- Thinking about maximum jump length only.
- Assuming sum of jumps matters.
Solution
Step 1: Check array declaration syntax in C#
In C#, arrays are declared with type[], then initialized with new type[] and values inside braces.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.Final Answer:
int[] jumps = new int[] {2, 3, 1, 1, 4}; -> Option CQuick Check:
C# array syntax = int[] jumps = new int[] {2, 3, 1, 1, 4}; [OK]
- Omitting 'new int[]' when initializing arrays.
- Using parentheses instead of braces for array values.
- Declaring arrays like in other languages (e.g., C/C++).
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}));Solution
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.Step 2: Understand the return value
Since index 4 is not reachable, the function returns false.Final Answer:
False -> Option DQuick Check:
Index 4 unreachable = False [OK]
- Assuming all arrays are reachable.
- Not updating maxReach correctly.
- Confusing true/false return values.
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;
}Solution
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.Step 2: Correct the loop boundary
The loop should run while i < nums.Length to avoid accessing beyond the array bounds.Final Answer:
The loop condition should be i < nums.Length, not i <= nums.Length. -> Option AQuick Check:
Array index out of range fixed by loop condition [OK]
- Using <= instead of < in loop conditions.
- Ignoring array index limits.
- Misunderstanding return types.
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;
}Solution
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.Step 2: Confirm algorithm correctness
The algorithm correctly identifies the last index is reachable without modification.Final Answer:
No modification needed; the current algorithm works correctly. -> Option BQuick Check:
Algorithm handles zeros and reachability correctly [OK]
- Thinking zeros always block progress.
- Adding unnecessary checks for zero values.
- Changing loop direction without reason.
Ready to prove it?
Pass the test and this pattern moves unknown
