The DP part clicked pretty fast, but I wasted a good few minutes just staring at the prime constraint.
Recognize this as a dynamic programming problem where dp[i] represents the maximum sum to reach index i. For each index, consider all valid previous indices from which you can jump to i (either 1 step back or x steps back where x is a prime ending in 3), and take the maximum dp value plus the current cell's value. Precompute all primes ending in 3 up to the array length to efficiently check valid jumps.
Pro tip: Mention that you can optimize space by only keeping track of the last x steps needed, but since x can be up to n, a full dp array is often simpler and still O(n * number of primes) time. Also, clarify edge cases like negative values and unreachable cells.
Ask about array size, possible values, and whether the last cell must be reached exactly. Confirm that you can only move forward and that x is a prime ending in 3.
Let dp[i] be the maximum sum to reach index i. Initialize dp[0] = arr[0]. For i > 0, dp[i] = arr[i] + max(dp[i-1], max over valid x of dp[i-x] if i-x >= 0).
Generate all primes up to n that end with digit 3 using sieve or trial division. Store them in a list for quick access during DP transitions.
Loop i from 1 to n-1, compute dp[i] using the recurrence, handling cases where no valid jump exists (set dp[i] to -infinity). Finally, return dp[n-1].
Time complexity is O(n * P) where P is the number of primes ending in 3 up to n. Space is O(n). Discuss possible optimizations like using a sliding window or segment tree if needed.
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