I think this method might fail when array contains multiple entries for same value like {1,2,2,2,4,6} and we requires Key=6.
then by following method given in video,
Number of pairs = 1
Original answer, Number of pairs = 3.
What change is needed to correctly solve above case ?
hello @karan_bhut
it will include all distinct pairs. some repetetion might be missing.
extract all unique numbers also store their freuency.
now apply the same two pointer logic . but this if if a[i]+a[j]==target
then print a[i] and a[j] pairs n1 * n2 times where n1 and n2 are the frequency of a[i] and a[j]
we have to check if pair is already considered or not right ? eg: 1,2,2,4,4,6 will give 8 pairs for key 6. but answer is 4 right ?
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u have to apply two pointers on unique numbers ,not on repeated array
Ohh okay understood.
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