if the ( a [ i ] + 1 ) is present then add that value as f ( a [ i ] , a [ i ] - 1 ) = 0 here we add as a [ i ] - ( a [ i ] - 1 ) < 0 which would have been added as negative sum , so we add to remove this pair from the sum value
if ( cnt [ a [ i ] + 1 ] ) ans += cnt [ a [ i ] + 1 ] ;