First find min {1 5 5}. In this case it's 1.

In general, you need to try the following targets: min, min-1, min-2, min-3 and min-4 and target >=0.

So in this particular case min=1 so we try only min and min-1.

Case 1: target is (1 1 1)

Find the delta between (1 5 5) and (1 1 1):

delta = (1 5 5) - (1 1 1) = (0 4 4)

In terms of {1, 2, 5} operations we can minimally partition (0 4 4) elements as follows:

0 = 0 * 5 + 0 * 2 + 0 * 1

4 = 0 * 5 + 2 * 2 + 0 * 1

4 = 0 * 5 + 2 * 2 + 0 * 1

Note that you get above partition as follows:

For some delta = (d0 d1 ... dn-1)

di = ai * 5 + bi * 2 + ci * 1

ai = di / 5

bi = (di % 5) / 2

ci = (((di % 5) % 2) / 1)

Now count the occurences of {1, 2, 5} in (0 4 4) partition and this is the min number of operations to reduce (1 5 5) -> (1 1 1):

min number of ops = (0 + 0 + 0) + (0 + 2 + 0) + (0 + 2 + 0) = 4

Case 2: target is (0 0 0)

Find the delta between (1 5 5) and (0 0 0):

delta = (1 5 5) - (0 0 0) = (1 5 5)

In terms of {1, 2, 5} operations we can minimally partition (1 5 5) elements as follows:

1 = 0 * 5 + 0 * 2 + 1 * 1

5 = 1 * 5 + 0 * 2 + 0 * 1

5 = 1 * 5 + 0 * 2 + 0 * 1

Now count the occurences of {1, 2, 5} in (1 5 5) partition and this is the min number of operations to reduce (1 5 5) -> (0 0 0):

min number of ops = (0 + 0 + 1) + (1 + 0 + 0) + (1 + 0 + 0) = 3

Case 1 requires 4 steps and case 2 requires 3 steps, so case 2 is optimal and min number of steps is 3.

Why do we not look min-5, min-6 etc? Suppose min-5 >=0 so (min-5 min-5 ... min-5) is a valid (non-negative elements) target.

delta' = (v0 - (min -5) ... vn-1 - (min -5)) = (v0 - min + 5 ... vn-1 - min + 5)

When we did (min ... min) target, delta was:

delta = (v0 - min ... vn-1 - min)

Now if you compare partitions of delta and delta' into {1, 2, 5} it's evident that delta' has one more 5 in each element di' so compared to number of ops in delta it requires n extra -5 operations. For that reason each of the cases min-5, min-6 etc are always less optimal than cases min, min-1, ... min-4.