##### ###### ##### ### # # ### # # ###### ## ## ## ## ## ## ## # # # # # ## ##### #### ##### # # # # # # # #### ## # ## ## ## ## # # # # # ## ## # ###### ## ### # ### # ######
##### ###### ##### ### # # ### # # ###### ## ## ## ## ## ## ## # # # # # ## ##### #### ##### # # # # # # # #### ## # ## ## ## ## # # # # # ## ## # ###### ## ### # ### # ######
The approach
Two passes: record last positions, then greedily partition.
def partitionLabels(s): last = {c: i for i, c in enumerate(s)} result = [] start = end = 0
for i, c in enumerate(s):
end = max(end, last[c])
if i == end:
result.append(end - start + 1)
start = i + 1
return result
time, space.