Jump Game Problem in DSA Rust - Time & Space Complexity
We want to understand how the time needed to solve the Jump Game problem changes as the input size grows.
Specifically, how many steps does the algorithm take to decide if we can reach the end?
Analyze the time complexity of the following code snippet.
fn can_jump(nums: Vec<i32>) -> bool {
let mut max_reach = 0;
for i in 0..nums.len() {
if i > max_reach {
return false;
}
max_reach = max_reach.max(i + nums[i] as usize);
}
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 position in the array once.
- How many times: The loop runs once for each element, so exactly n times where n is the array length.
As the input size grows, the number of steps grows directly with it.
| 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 input size increases.
Time Complexity: O(n)
This means the time to decide if you can reach the end grows directly with the number of positions.
[X] Wrong: "Because we check jumps from each position, the time must be much bigger than n, maybe n squared."
[OK] Correct: The loop only goes through the array once, and we keep track of the farthest reachable position without extra nested loops.
Understanding this linear time solution shows you can find efficient ways to solve problems by tracking progress smartly, a key skill in interviews.
"What if we changed the problem to find the minimum number of jumps needed? How would the time complexity change?"
Practice
Solution
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.Step 2: Identify the goal
The goal is to check if the last index is reachable, not the number of jumps or sums.Final Answer:
If you can reach the last index starting from the first index -> Option CQuick Check:
Jump Game = Reach last index [OK]
- Confusing reachability with minimum jumps
- Thinking about total jump sums
- Assuming jumps can be backward
Solution
Step 1: Identify Rust vector syntax
In Rust, vectors are declared using vec![elements], which creates a growable array.Step 2: Check each option
let jumps = vec![2, 3, 1, 1, 4]; uses vec![], which is correct. let jumps = [2, 3, 1, 1, 4]; is an array, not a vector. let jumps = (2, 3, 1, 1, 4); is a tuple. let jumps = {2, 3, 1, 1, 4}; uses braces which is invalid syntax.Final Answer:
let jumps = vec![2, 3, 1, 1, 4]; -> Option AQuick Check:
vec! macro creates vectors [OK]
- Using array syntax instead of vec! macro
- Confusing tuples with vectors
- Using braces {} which is invalid
fn can_jump(nums: Vec<usize>) -> bool {
let mut max_reach = 0;
for i in 0..nums.len() {
if i > max_reach {
return false;
}
max_reach = max_reach.max(i + nums[i]);
}
true
}
fn main() {
let jumps = vec![2, 3, 1, 1, 4];
println!("{}", can_jump(jumps));
}Solution
Step 1: Trace the loop and max_reach updates
Start with max_reach = 0. For i=0, max_reach = max(0, 0+2) = 2. For i=1, max_reach = max(2, 1+3) = 4. For i=2, max_reach = max(4, 2+1) = 4. For i=3, max_reach = max(4, 3+1) = 4. For i=4, max_reach = max(4, 4+4) = 8.Step 2: Check if any index is unreachable
At each i, i <= max_reach, so no false return. Loop completes and returns true.Final Answer:
true -> Option DQuick Check:
Max reach covers all indices [OK]
- Not updating max_reach correctly
- Returning false too early
- Confusing max_reach with jump count
fn can_jump(nums: Vec<usize>) -> bool {
let mut max_reach = 0;
for i in 0..nums.len() {
if i >= max_reach {
return false;
}
max_reach = max_reach.max(i + nums[i]);
}
true
}Solution
Step 1: Analyze the condition for unreachable index
The condition 'if i >= max_reach' returns false even when i equals max_reach, which is incorrect because reaching exactly max_reach is allowed.Step 2: Correct the condition
Changing it to 'if i > max_reach' allows reaching max_reach index and only returns false if i is beyond max_reach.Final Answer:
Change 'if i >= max_reach' to 'if i > max_reach' -> Option BQuick Check:
Allow index equal to max_reach [OK]
- Using >= instead of > causes early false
- Using min instead of max for max_reach
- Skipping index 0 in loop
vec![3, 2, 1, 0, 4], what is the correct output of the Jump Game problem using the standard greedy approach in Rust?Solution
Step 1: Trace max_reach for each index
Start max_reach=0. i=0: max_reach=max(0,0+3)=3. i=1: max_reach=max(3,1+2)=3. i=2: max_reach=max(3,2+1)=3. i=3: max_reach=max(3,3+0)=3. i=4: i=4 > max_reach=3, so return false.Step 2: Understand why last index is unreachable
Index 4 cannot be reached because max_reach never extends beyond 3 due to zero jump at index 3.Final Answer:
false -> Option AQuick Check:
Zero jump blocks progress [OK]
- Assuming always true if first jump > 0
- Ignoring zero jump blocking path
- Confusing max_reach with jump count
Ready to prove it?
Pass the test and this pattern moves unknown
