Demo preview: These solutions pass the automated test suite but have not been human-reviewed. Treat trade-offs and prose as draft.

Mediumgreedy

Jump Game II

Problem (restated)

Minimum jumps to reach last index (guaranteed reachable).

Intuition

Level-by-level: within current jump range, track farthest next end; when i hits end, jump++.

Approaches

Range BFS greedy

Unverified

Time O(n)Space O(1)

Idea. curEnd, far; when i==curEnd, jumps++, curEnd=far.

Walkthrough. [2,3,1,1,4] → 2 jumps.

Trade-offs. Greedy levels vs DP min jumps O(n²).

Solution

export function jump(nums: number[]): number {
  let jumps = 0, curEnd = 0, far = 0;
  for (let i = 0; i < nums.length - 1; i++) {
    far = Math.max(far, i + nums[i]!);
    if (i === curEnd) {
      jumps++;
      curEnd = far;
    }
  }
  return jumps;
}

export function jump(nums: number[]): number {
  let jumps = 0, curEnd = 0, far = 0;
  for (let i = 0; i < nums.length - 1; i++) {
    far = Math.max(far, i + nums[i]!);
    if (i === curEnd) {
      jumps++;
      curEnd = far;
    }
  }
  return jumps;
}

Template connection

Greedy jump with range expansion.

Reflection

The farthest jump inside the current window is the next window. The number of windows is the number of jumps.
Stop once the last index is already inside this window. [0] answers 0. A jump of 1 at every index is n - 1.
The windows are BFS layers. The greedy choice is the max reach inside the window.