Approach #1: Greedy [Accepted]
Intuition
Let's try to repeatedly choose the smallest leftjustified partition.
Consider the first label, say it's 'a'
. The first partition must include it, and also the last occurrence of 'a'
.
However, between those two occurrences of 'a'
, there could be other labels that make the minimum size of this partition bigger. For example, in "abccaddbeffe"
, the minimum first partition is "abccaddb"
.
This gives us the idea for the algorithm: For each letter encountered, process the last occurrence of that letter, extending the current partition [anchor, j]
appropriately.
Algorithm
We need an array last[char] > index of S where char occurs last
.
Then, let anchor
and j
be the start and end of the current partition.
If we are at a label that occurs last at some index after j
, we'll extend the partition j = last[c]
. If we are at the end of the partition (i == j
) then we'll append a partition size to our answer, and set the start of our new partition to i+1
.
Complexity Analysis

Time Complexity: , where is the length of .

Space Complexity: .
Analysis written by: @awice.