55. Jump Game
Problem Description
You are given an array nums
, where each element represents the maximum number of steps you can jump forward from that position. Your goal is to determine if you can reach the last element of the array starting from the first element. You start at the first index of the array, and at each step, you can jump forward to any position within your current element's range. If you can make it to any index that has enough range to reach the end, or if the end is within your reach from where you are, the answer is true
. Otherwise, it's false
.
Intuition
The key intuition behind the solution is to take a greedy approach. This means we want to make the locally optimal choice at each step in the hope that this will lead to the globally optimal solution. In this case, as we traverse the array, we continuously keep track of the furthest we can reach (mx
).
We start at the first index and iterate through the array. For each index, we update mx
to be the maximum of its current value or the furthest we can get from this index (which is the current index i
plus the jump length nums[i]
). As we do this, if we find mx
is ever less than i
, we cannot reach position i
or beyond, which means we cannot reach the last index, and return false
. If we can always reach the position we're iterating over, by the end of the iteration, we can reach the end, and we return true
.
Learn more about Greedy and Dynamic Programming patterns.
Solution Approach
The solution uses a simple but effective greedy algorithm. The implementation of this solution does not require any complex data structures, relying only on a single integer variable mx
which keeps track of the maximum index that can be reached at each step.
The following steps summarize the implementation of the solution:
- We initialize a variable
mx
to 0.mx
represents the maximum index that can be reached so far. - We iterate through each index
i
of thenums
array using afor
loop. Theenumerate
function is handy here as it gives us both the indexi
and the valuex
at that index. - During each iteration, we first check if
mx
is less thani
. If this is the case, we returnfalse
because it indicates that we can't reach the current index (or any index beyond it). - If
mx
is not less thani
, it means we can reach this position, and therefore we updatemx
to be the maximum of its current value andi + x
, wherex
is the maximum jump length from the current position. The expressionmax(mx, i + x)
efficiently accomplishes this. - After the loop, if we never encountered a situation where
mx < i
, we have been able to reach or surpass each index, including the last index. Thus, we returntrue
.
The code or mathematical formulas are encapsulated using backticks, for example, mx = max(mx, i + nums[i])
.
By always keeping track of the furthest reachable index and bailing out early if we find an unreachable index (mx < i
), we ensure an efficient solution with a time complexity that is linear with respect to the length of the nums
array.
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Start EvaluatorExample Walkthrough
Let's walk through a small example using the greedy solution approach outlined above. Suppose we are given the following array nums
:
nums = [2, 3, 1, 1, 4]
Our task is to determine if we can reach the last element, which in this case is 4
. We will use the steps of the solution to illustrate how we can arrive at the answer:
-
Initialize
mx
to0
. This means before we start, the furthest index we can reach is0
. -
Start iterating:
- At index
0
, the value is2
. We can jump to index1
or2
. Updatemx
tomax(0, 0 + 2)
which is2
. - Move to index
1
, with a value of3
. Now we can reach as far as1 + 3 = 4
, which is the end of the array. Updatemx
tomax(2, 1 + 3)
which is4
. - At this point, we can already reach the last element, so we could stop and return
true
. Nonetheless, for completeness: - At index
2
, the value is1
. From here,max(4, 2 + 1)
is still4
. - At index
3
, the value is1
. From here,max(4, 3 + 1)
is still4
.
- At index
-
Throughout the iteration,
mx
was always greater than or equal toi
, so we never returnedfalse
. -
After the loop, since there were no indices that we could not reach, and we were always able to update
mx
to reach or go beyond the next index, we can returntrue
.
So, in our example, it is indeed possible to reach the last element starting from the first element following the greedy approach.
Solution Implementation
1from typing import List
2
3class Solution:
4 def canJump(self, nums: List[int]) -> bool:
5 # Initialize the variable `max_reach` which represents the maximum
6 # index we can reach so far.
7 max_reach = 0
8
9 # Enumerate through the list, with `index` representing the current
10 # position and `jump_length` the maximum jump length from that position.
11 for index, jump_length in enumerate(nums):
12 # If our current `max_reach` is less than the `index`, we can't
13 # jump to `index` or beyond, hence we return False.
14 if max_reach < index:
15 return False
16
17 # Update `max_reach` by the farthest we can get from here, which
18 # is either the current `max_reach` or `index + jump_length`.
19 max_reach = max(max_reach, index + jump_length)
20
21 # If we've gone through all elements without returning False, it means
22 # we can reach the end of the list, so we return True.
23 return True
24
1class Solution {
2 public boolean canJump(int[] nums) {
3 int maxReachable = 0; // Initialize the maximum reachable index to 0
4
5 // Iterate over each index in the array
6 for (int i = 0; i < nums.length; ++i) {
7 // If the current index is greater than the maximum reachable index,
8 // it means we cannot proceed further, so return false.
9 if (maxReachable < i) {
10 return false;
11 }
12
13 // Update the maximum reachable index if the reachable index
14 // from the current position is greater than the previous max.
15 maxReachable = Math.max(maxReachable, i + nums[i]);
16 }
17
18 // If the loop completes without returning false, it means we can
19 // reach the last index, so return true.
20 return true;
21 }
22}
23
1#include <vector> // Include the necessary header for vector
2using namespace std;
3
4class Solution {
5public:
6 // Function to check if we can jump to the last index of the vector 'nums'
7 bool canJump(vector<int>& nums) {
8 int maxReachable = 0; // Variable to keep track of the maximum reachable index so far
9
10 // Iterate through each element of the vector
11 for (int i = 0; i < nums.size(); ++i) {
12 // If the current index is greater than the maximum reachable index, we can't proceed, return false
13 if (maxReachable < i) {
14 return false;
15 }
16
17 // Update maxReachable to the maximum of the current maxReachable and the current index plus its jump length
18 maxReachable = max(maxReachable, i + nums[i]);
19 }
20
21 // If we are able to iterate through the entire vector, return true
22 return true;
23 }
24};
25
1// Determines if it is possible to reach the last index of the array
2// nums[i] represents the maximum jump length from that position.
3function canJump(nums: number[]): boolean {
4 let maxReach: number = 0; // Variable to keep track of the maximum reachable index
5 for (let currentIndex = 0; currentIndex < nums.length; ++currentIndex) {
6 // Check if the current index is beyond the maximum reachable index
7 if (maxReach < currentIndex) {
8 // If we cannot reach this index, return false
9 return false;
10 }
11 // Update the maximum reachable index
12 maxReach = Math.max(maxReach, currentIndex + nums[currentIndex]);
13 }
14 // If we can reach beyond the last index, return true
15 return true;
16}
17
Time and Space Complexity
The given Python code aims to determine whether it is possible to jump to the last index of the given list nums
. The function canJump
works by iterating through each element in the list, calculating the maximum distance that can be reached from the current position, and checking whether that distance is sufficient to continue progressing through the array.
-
Time Complexity: The time complexity of the code is
O(n)
, wheren
is the length of the arraynums
. This is because the function involves a single loop that goes through the array once, making a constant-time check and update at each step. -
Space Complexity: The space complexity of the code is
O(1)
. The algorithm uses a fixed amount of additional space (the variablemx
), regardless of the input size, so the space used does not grow with the size of the input array.
Learn more about how to find time and space complexity quickly using problem constraints.
Consider the classic dynamic programming of longest increasing subsequence:
Find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.
For example, the length of LIS for [50, 3, 10, 7, 40, 80]
is 4
and LIS is
[3, 7, 40, 80]
.
What is the recurrence relation?
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