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Java-based LeetCode algorithm problem solutions, regularly updated
package g0001_0100.s0055_jump_game;
// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Dynamic_Programming #Greedy
// #Algorithm_II_Day_12_Dynamic_Programming #Dynamic_Programming_I_Day_4 #Udemy_Arrays
// #2023_08_11_Time_2_ms_(79.47%)_Space_44.8_MB_(22.14%)
public class Solution {
public boolean canJump(int[] nums) {
int sz = nums.length;
// we set 1 so it won't break on the first iteration
int tmp = 1;
for (int i = 0; i < sz; i++) {
// we always deduct tmp for every iteration
tmp--;
if (tmp < 0) {
// if from previous iteration tmp is already 0, it will be <0 here
// leading to false value
return false;
}
// we get the maximum value because this value is supposed
// to be our iterator, if both values are 0, then the next
// iteration we will return false
// if either both or one of them are not 0 then we will keep doing this and check.
// We can stop the whole iteration with this condition. without this condition the code
// runs in 2ms 79.6%, adding this condition improves the performance into 1ms 100%
// because if the test case jump value is quite large, instead of just iterate, we can
// just check using this condition
// example: [10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] -> we can just jump to the end without
// iterating whole array
tmp = Math.max(tmp, nums[i]);
if (i + tmp >= sz - 1) {
return true;
}
}
// we can just return true at the end, because if tmp is 0 on previous
// iteration,
// even though the next iteration index is the last one, it will return false under the
// tmp<0 condition
return true;
}
}