Constrained Sum - Walkthrough

Find maximum sum of subsequence where adjacent elements are at most $k$ apart. $dp[i]$ = max sum ending at $i$. Transition: $dp[i] = nums[i] + \max(0, \max_{j \in [i-k, i-1]} dp[j])$. The $\max(0, ...)$ handles starting fresh. Example: $[10, 2, -10, 5, 20]$, $k = 2$. $dp = [10, 12, 2, 17, 37]$. At each step, query window max and add current.

Answer: $\max(dp) = 37$ (subsequence 10, 2, 5, 20). Monotonic queue handles the window max in $O(1)$ amortized.