Select i and j starts from i+1, pick the minimum element while completing the j loop, and then swap it with i, then again continue like this
"Select the minimum element, then swap with i"
Heaviest Element pulled towards the end.
outer loop starts from the end, inner j loop from 0 to i-1
we maintain a sorted left array as we move forward, so for i we gonna start from there and go till index 0 (j), make sure that this subarray array is sorted.
continue until we reach the end, that means from end to index 0, the whole array becomes sorted.
Consider the frequency as index.
make a map that stores elements and its frequencies
then a vector/ list of list that stores the elements, based on frequency as index and list as the elements that have frequency == that array index
ABSOLUTE CINEMA!