Jump Game I, a linear-time, constant-space approach
Problem is here: https://leetcode.com/problems/jump-game/
55. Jump Game
One approach to achieve linear-time and constant-space would be:
- Keep track of the current jump length (jump = nums[0])
- Iterate over all the elements from index one onward
- If the jump length ever becomes zero or -ve, return false (can't move fwd)
- The new jump length becomes the maximum between the current element and jump length -1
Code is below, cheers, ACC.
public class Solution
{
public bool CanJump(int[] nums)
{
int n = nums[0];
for (int i = 1; i < nums.Length; i++)
{
if (n <= 0) return false;
n = Math.Max(n - 1, nums[i]);
}
return true;
}
}
55. Jump Game
Medium
Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Determine if you are able to reach the last index.
Example 1:
Input: [2,3,1,1,4] Output: true Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: [3,2,1,0,4] Output: false Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
Accepted
342,775
Submissions
1,039,505
One approach to achieve linear-time and constant-space would be:
- Keep track of the current jump length (jump = nums[0])
- Iterate over all the elements from index one onward
- If the jump length ever becomes zero or -ve, return false (can't move fwd)
- The new jump length becomes the maximum between the current element and jump length -1
Code is below, cheers, ACC.
public class Solution
{
public bool CanJump(int[] nums)
{
int n = nums[0];
for (int i = 1; i < nums.Length; i++)
{
if (n <= 0) return false;
n = Math.Max(n - 1, nums[i]);
}
return true;
}
}
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