Cut an array into contiguous groups, each no longer than `k`, then replace every element of a group with that group's own largest value. Choose the grouping that makes the total, after every replacement, as large as possible. Once a group's length is fixed, its contribution is fixed too — the group's maximum, counted once per element in it — so the only real decision at each position is how long the next group should run, from 1 up to k. Reaching some position i for the first time as the start of a fresh group means an earlier position was already settled optimally, and the stretch between them is exactly one new group. So the best total ending at i is the best total ending at some valid earlier cut, plus that new group's contribution — and trying every legal group length at every position, building left to right, means every earlier answer needed is already on hand.