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Java-based LeetCode algorithm problem solutions, regularly updated

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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
// #Big_O_Time_O(n)_Space_O(1) #2023_08_11_Time_2_ms_(79.47%)_Space_44.8_MB_(22.14%)

/**
 * 55 - Jump Game\.
 *
 * Medium
 *
 * You are given an integer array `nums`. You are initially positioned at the array's **first index** , and each element in the array represents your maximum jump length at that position.
 *
 * Return `true` _if you can reach the last index, or_ `false` _otherwise_.
 *
 * **Example 1:**
 *
 * **Input:** nums = [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:** nums = [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. 
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁴
 * *   0 <= nums[i] <= 10⁵
**/
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;
    }
}