Jump Game I and II.

Problem. Given an array nums where nums[i] is the maximum jump length from index i:

• Jump Game I — can you reach the last index? Return a boolean.
• Jump Game II — return the minimum number of jumps to reach the last index (assume it's always reachable).

Jump Game I — track the farthest reachable index

The greedy insight: you don't need to know how you got somewhere, only the farthest index reachable so far. Sweep left to right; if the current index is ever beyond what's reachable, you're stuck.

boolean canJump(int[] nums) {
    int farthest = 0;
    for (int i = 0; i < nums.length; i++) {
        if (i > farthest) return false;          // can't even reach i
        farthest = Math.max(farthest, i + nums[i]);
        if (farthest >= nums.length - 1) return true;
    }
    return true;
}

O(n) time, O(1) space. The greedy choice — always extend farthest — is safe because reachability is monotone: if index i is reachable, so is every index up to farthest.

Jump Game II — implicit BFS by levels

Minimizing jumps is a shortest-path-by-edges problem, so it's BFS — but you can run it in O(n) without a queue. Treat each "jump" as a BFS level: [curEnd] is the farthest index reachable with the current jump count. When you reach the end of a level, you must spend one more jump, and the new boundary is the farthest you saw in this level.

int jump(int[] nums) {
    int jumps = 0, curEnd = 0, farthest = 0;
    for (int i = 0; i < nums.length - 1; i++) {   // stop before last index
        farthest = Math.max(farthest, i + nums[i]);
        if (i == curEnd) {                        // exhausted current jump's range
            jumps++;
            curEnd = farthest;                    // commit to the best reachable boundary
        }
    }
    return jumps;
}

We loop to length - 1 so we don't count a phantom jump from the last index. Each time i hits curEnd, we've consumed one BFS level and extend the boundary to farthest — the greedy choice that this level's best reach is the optimal next boundary.