@amol..we can simply break the loop where we find the slow and fast pointer meeting,then its become a a problem of two link list having common ending point and different staring point and for detecting the extact point where loop began is two traverse both the list and find its length lets say 'a' and 'b' now put 1st pointer at start position of 1st link list and other pointer at distance of |a-b| from second list ...now where both of them meet is the point where loop began.
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