Can someone explain possibly with an example, how we assumed or concluded i=0∼m, j=(m+n+1)/2 − i ?
I got how i+j=m−i+n−j (or: m - i + n - j + 1m−i+n−j+1) part but couldn't get my head around j = (m+n+1)/2 − i.
Posts made by kool
RE: Sum of Count of Different bits
It's actually similar to finding total hamming distance.