Solving the Jump Game II Problem in JavaScript

 

Introduction

When faced with algorithmic challenges, one of the most rewarding aspects is finding an efficient solution to a seemingly complex problem. One such challenge is the Jump Game II problem on LeetCode, which asks us to determine the minimum number of jumps required to reach the last index of an array. In this blog, we'll walk through the problem, discuss the approach, and dive into the JavaScript implementation.

Problem Statement

You are given an array of non-negative integers where each element represents the maximum number of steps you can jump forward from that position. Your task is to find the minimum number of jumps needed to reach the last index, starting from the first index.

Approach

The problem requires us to identify the minimum jumps to reach the end of the array. A brute force solution would involve exploring all possible jump paths, but that would be inefficient. Instead, a greedy approach proves to be optimal.

Key Ideas:

  1. Start at the First Index: Begin at the first index of the array and decide the next index to jump to.
  2. Evaluate Possible Jumps: From each index, evaluate all possible jumps within the range specified by the value at that index.
  3. Select the Optimal Jump: Choose the jump that allows you to reach the furthest possible point.
  4. Increment the Jump Count: Each time you make a jump, increment your count. Continue this until you reach the last index.

Time complexity: O(n)

Space complexity: O(1)

Code Implementation

Let’s break down the JavaScript implementation of this approach:

/**
 * @param {number[]} nums
 * @return {number}
 */
var jump = function(nums) {
    let count = 0;
    let nextIndex;
    let i = 0;
    while(i !== nums.length - 1){
        if( i + nums[i] >= nums.length-1){
            return count + 1;
        }
        let arrayOfChoices = nums.slice(i+1, i + nums[i] + 1);
        let maxJump = 0;
        for(let j = 0; j < arrayOfChoices.length; j++){
            if(j + arrayOfChoices[j] >= maxJump) {
                maxJump = j + arrayOfChoices[j];
                nextIndex = i + j + 1;
            }
        }
        i = nextIndex;
        count++;
    }
    return count;
};

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