Jump Game II — LeetCode #45

You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i] and
• i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

Example 1:

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: nums = [2,3,0,1,4]
Output: 2

Constraints:

• 1 <= nums.length <= 104
• 0 <= nums[i] <= 1000

Solutions:

Python

class Solution:
    def jump(self, nums: List[int]) -> int:
        n = len(nums)
        if n < 2:
            return 0
        max_reach, steps, end = 0, 0, 0
        for i in range(n - 1):
            max_reach = max(max_reach, i + nums[i])
            if i == end:
                end = max_reach
                steps += 1
        return steps

C#

public class Solution {
    public int Jump(int[] nums) {
        int n = nums.Length;
        if (n < 2) {
            return 0;
        }
        int maxReach = 0, steps = 0, end = 0;
        for (int i = 0; i < n - 1; i++) {
            maxReach = Math.Max(maxReach, i + nums[i]);
            if (i == end) {
                end = maxReach;
                steps++;
            }
        }
        return steps;
    }
}

Java

class Solution {
    public int jump(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return 0;
        }
        int maxReach = 0, steps = 0, end = 0;
        for (int i = 0; i < n - 1; i++) {
            maxReach = Math.max(maxReach, i + nums[i]);
            if (i == end) {
                end = maxReach;
                steps++;
            }
        }
        return steps;
    }
}

Javascript

/**
 * @param {number[]} nums
 * @return {number}
 */
var jump = function(nums) {
    let n = nums.length;
    if (n < 2) {
        return 0;
    }
    let maxReach = 0, steps = 0, end = 0;
    for (let i = 0; i < n - 1; i++) {
        maxReach = Math.max(maxReach, i + nums[i]);
        if (i === end) {
            end = maxReach;
            steps++;
        }
    }
    return steps;
};

Typescript

function jump(nums: number[]): number {
    let n = nums.length;
    if (n < 2) {
        return 0;
    }
    let maxReach = 0, steps = 0, end = 0;
    for (let i = 0; i < n - 1; i++) {
        maxReach = Math.max(maxReach, i + nums[i]);
        if (i === end) {
            end = maxReach;
            steps++;
        }
    }
    return steps;
}

PHP

class Solution {

    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function jump($nums) {
        $n = count($nums);
        if ($n < 2) {
            return 0;
        }
        $maxReach = 0;
        $steps = 0;
        $end = 0;
        for ($i = 0; $i < $n - 1; $i++) {
            $maxReach = max($maxReach, $i + $nums[$i]);
            if ($i === $end) {
                $end = $maxReach;
                $steps++;
            }
        }
        return $steps;
    }
}

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