 Solution

Let $D[j]$ be the minimum number of jumps from index zero to index $j$, with $\infty$ denoting an unreachable position. The minimum forward-jump dynamic programming recurrence iswhere the minimum of an empty set is $\infty$ and $1+\infty =\infty$. Compute it in increasing index order:

D[0] = 0
for j = 1, ..., n-1:
    D[j] = INFINITY
    for i = 0, ..., j-1:
        if D[i] != INFINITY and A[i] >= j-i:
            D[j] = min(D[j], D[i] + 1)
return D[n-1]

For algorithm correctness, classify any route to $j>0$ by its penultimate index $i$. Every jump advances, so $i<j$, and legality requires $A[i]\ge j-i$. By the principle of optimality, a shortest prefix to $i$ can be used; this gives the lower bound represented by the recurrence. Conversely, a minimizing reachable predecessor supplies an actual route of length $D[i]+1$, proving attainability. Mathematical induction on $j$ therefore proves every stored value correct.

There are $n(n-1)/2$ predecessor tests, and two arrays are unnecessary when only the number is required. ThusFor the source example, the computed array isWhen $n=1$, zero jumps are needed even if the lone entry is zero. An unreachable final position returns $\infty$. Testing A[i] >= j-i also avoids forming a potentially overflowing sum i + A[i] in bounded implementations. A faster frontier-based algorithm is possible, but the displayed method directly supplies the requested dynamic programming solution.