g0001_0100.s0055_jump_game.Solution.scala Maven / Gradle / Ivy

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104 LeetCode algorithm problem solutions

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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) #2024_06_01_Time_769_ms_(79.03%)_Space_59.8_MB_(50.00%)

object Solution {
    def canJump(nums: Array[Int]): Boolean = {
        val sz = nums.length
        // We set tmp to 1 so it won't break on the first iteration
        var tmp = 1
        var result = true // Variable to store the result

        for (i <- 0 until sz if result) {
            // We always deduct tmp for every iteration
            tmp -= 1
            if (tmp < 0) {
                // If from the previous iteration tmp is already 0, it will be < 0 here,
                // leading to a false value
                result = false
            } else {
                // We get the maximum value because this value is supposed to be our iterator. If both values are 0,
                // then the next iteration will return false.
                // 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 to 1ms (100%)
                // because if the test case jump value is quite large, instead of just iterating, 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 the whole array.
                tmp = math.max(tmp, nums(i))
                if (i + tmp >= sz - 1) {
                    result = true
                }
            }
        }

        result
    }
}